Summary Of Trigonometric Identities 2z2d2m

Summary of trigonometric identities You have seen quite a few trigonometric identities in the past few pages. It is convenient to have a summary of them for reference. These identities mostly refer to one angle denoted t, but there are a few of them involving two angles, and for those, the other angle is denoted s.. More important identities

You don't have to know all the identities off the top of your head. But these you should. Defining relations for tangent, cotangent, secant, and cosecant in of sine and cosine. tan t = sec t =

sin t cos t 1 cos t

cot t = csc t =

1 tan t 1

=

cos t sin t

sin t

The Pythagorean formula for sines and cosines. sin2 t + cos2 t = 1 Identities expressing trig functions in of their complements cos t = sin( /2 – t)

sin t = cos( /2 – t)

cot t = tan( /2 – t)

tan t = cot( /2 – t)

csc t = sec( /2 – t)

sec t = csc( /2 – t)

Periodicity of trig functions. Sine, cosine, secant, and cosecant have period 2 while tangent and cotangent have period . sin (t + 2 ) = sin t cos (t + 2 ) = cos t tan (t + ) = tan t Identities for negative angles. Sine, tangent, cotangent, and cosecant are odd functions while cosine and secant are even functions. sin –t = –sin t cos –t = cos t tan –t = –tan t Sum formulas for sine and cosine sin (s + t) = sin s cos t + cos s sin t cos (s + t) = cos s cos t – sin s sin t Double angle formulas for sine and cosine sin 2t = 2 sin t cos t cos 2t = cos2 t – sin2 t = 2 cos2 t – 1 = 1 – 2 sin2 t Less important identities

You should know that there are these identities, but they are not as important as those mentioned above. They can all be derived from those above, but sometimes it takes a bit of work to do so.

The Pythagorean formula for tangents and secants. sec2 t = 1 + tan2 t Identities expressing trig functions in of their supplements sin( – t) = sin t cos( – t) = –cos t tan( – t) = –tan t Difference formulas for sine and cosine sin (s – t) = sin s cos t – cos s sin t cos (s – t) = cos s cos t + sin s sin t Sum, difference, and double angle formulas for tangent tan (s + t) = tan (s – t) = tan 2t =

tan s + tan t 1 – tan s tan t tan s – tan t 1 + tan s tan t 2 tan t 1 – tan2 t

Half-angle formulas sin t/2 = ± ((1 – cos t) / 2) cos t/2 = ± ((1 + cos t) / 2) tan t/2 =

sin t 1 + cos t

=

1 – cos t sin t

Truly obscure identities

These are just here for perversity. Yes, of course, they have some applications, but they're usually narrow applications, and they could just as well be forgotten until, if ever, needed. Product-sum identities sin s + sin t = 2 sin sin s – sin t = 2 cos cos s + cos t = 2 cos cos s – cos t = –2 sin

s+t 2 s+t 2 s+t 2 s+t 2

cos sin cos sin

s–t 2 s–t 2 s–t 2 s–t 2

Product identities sin s cos t = sin (s + t) + sin (s – t)

Aside: weirdly enough, these product identities were used before logarithms to perform multiplication. Here's how you could use the second one. If you want to multiply xtimes y, use a table to look up the angle s whose cosine is x and the angle t whose cosine is y. Look up the cosines of the sum s + t, and the

2 cos (s + t) + cos (s – t) cos s cos t =

difference s – t. Average those two cosines. You get the product xy! Three table look-ups, and computing a sum, a difference, and an average rather than one multiplication. Tycho Brahe (1546-1601), among others, used this algorithm known as prosthaphaeresis.

2 cos (s – t) – cos (s + t ) sin s sin t = 2

Triple angle formulas. You can easily reconstruct these from the addition and double angle formulas. sin 3t = 3 sin t – 4 sin3 t cos 3t = 4 cos3 t –3 cos t tan 3t =

3 tan t – tan3t 1 – 3 tan2t

More half-angle formulas. (These are used in calculus for a particular kind of substitution in integrals sometimes called the Weierstrass t-substitution.) sin t =

2 tan t/2 1 + tan2 t/2

cos t =

1 – tan2 t/2

tan t =

1 + tan2 t/2

2 tan t/2 1 – tan2 t/2

A Table of Exact Trig values that are expressible as simple involving square-roots. a

b

rad deg ian ree s s

0

0

cos(a) sin(b)

tan(a) cot(b)

1

0

90

√6 – √3 + √2 – 2 (√2 – 1)(√3 – √2)

82. 5

2 – √3 tan(15°) = [0; 3, [1, 2]] tan2(15°) = [0; 13, [1, 12]]

75

5π 12

72

2π 5

π 7.5

24

√6

π 15

4

12

√2 +

1 4

=

√ 3 +

4 =

+

2 4 cos2(15°) = [0; 1, 13, [1, 12]]

π 18

5

5

√ 12

8

8

1

5

–

10

4

8 2 cos (18°) = [0; 1, 9, [2, 8]]

8

deg rad ree ian s s

π 2 11

π 24

2√5 5 tan2(18°) = [0; 9, [2,8]]

2 + √2

π 22· 8

√ 8

4

5

=

2

+ 8 cos2(22·5°) = [0; 1, 5, [1, 4]]

8

3

π 30

=

6

√2 – 1 tan(22·5) = [0; [2]] tan2(22·5) = [0; 5, [1,4]]

4

√3

√3

2

3

cos(30°) = [0; 1, 6, [2, 6]] cos2(30°) = [0; 1, 3] √5

π 36

1 +

4

5

4

3

π 3

√5

=

+

8

8

√ 5 – 2√5 tan2(36°) = [0; 1, 1, [8, 2]]

5π 37. 24 5

√6 + √3 – √2 – 2 (√2 + 1)(√3 – √2)

54

3π 10

52. 7π 5 24

1

2 =

4

60

tan(30°) = [0; 1, [1,2]] tan2(30°) = [0; 3]

cos(36°) = [0; 1, [4]] cos2(36°) = [0; 1, 1, 1, [8, 2]]

π 45

67· 3π 5 8

√2

4

1

45

√6 – √3 – √2 + 2 (√2 – 1)(√3 + √2)

52. 5

π 2

cos(45°) = [0; 1, [2]] cos2(45°) = [0; 2] 7π 52. 24 5 1√ 05 3π 54 10

5

=

π 24

√ 5 –

–

5

1

8

8

2

+

2√5 5

36

π

30

π

5

tan2(54°) = [1; 1, [8, 2]] 4 cos2(54°) = [0; 2, 1, [8, 2]]

π 60 3

1

1 = 4

2

√3 tan(60°) = [1; [1,2]] tan2(60°) = 3

6

cos(60°) = [0; 2] cos2(60°) = [0; 4] 2 – √2

√ 8

4

3π 67· 8 5

=

2

2π 72 5

– 8 2 cos (67·5) = [0; 6, [1, 4]]

√5

1 –

4

1 + √2 tan(67·5) = [2; [2]] tan2(67·5) = [5; [1,4]]

8

3

√5

=

4

–

8

√ 5 + 2√5

8

5π 75 12

4

√2 –

1 =

4

√ 3 –

4 =

2 4 2 cos (75°) = [0; 14, [1, 12]] 11

π 24

√ 12 –

8

8

82. 5

π 90

0

2

18 tan2(72°) = [9; [2, 8]]

cos(72°) = [0; 3, [4]] cos2(72°) = [0; 10, [2, 8] ] √6

22· 5

2 + √3 tan(75°) = [3; [1, 2]] tan2(75°) = [13; [1,12]]

15

√6 + √3 + √2 + 2 (√2 + 1)(√3 + √2)

7.5

Infinity

0

π 8

π 10

π 12

π 24 0

The values in the table are those angles of the form n° or n/2 or n/3 for a whole number n, between 0 and 90° whose sin or cosine is rational, or whose continued fraction is periodic or the square of the trig value has a periodic continued fraction. 1

Continued fraction [a; b,c,d,...] means a + 1 b+ 1 c+ d + ...

and the periodic continued fraction [a;b, c, d, e, d, e, d, e, d, e,...] is written as [a; b, c, [d, e]].

Trig functions of Angles <0 or >90° Top of Form

To find the trig. values of all angles including those bigger than 90 degrees and negative angles: 1. select a trig function 2. type the angle in the box and then 3.

click on the button

to find which angle in the range 0-90° has the same value: Select a trig function:

(

°)

Degrees-Radians Converter To convert between DEGREES and RADIANS: 1. enter the angle as a number in one box leaving the other empty 2.

then click the button to do the conversion

You can use Pi in the radians box and * for multiplication e.g. 3*Pi/2: degrees

radians

Bottom of Form

Patterns The Simple Square-Root pattern Ernesto La Orden of Madrid pointed out the following neat way to connect and the easiest of the sines (cosines): Angl e

sine cosine √4

90

30

0

= 1

2

0

√3

60

45

Angl e

30

2 √2

1

=

2

√2

√1

1

=

2

2 √0 2

=0

45

60

90

The √(2 ± Phi) pattern 1 2

cos( = 9°)

2 + 2+Φ cos( = 18°)

1 2

2 2+φ +

1 =

2

2+Φ

1 2

cos( = 27°)

2

cos( = 36°)

1 2

2 2–Φ +

+ 2–φ 1 =

2

2+φ

Φ =

2

1

1

2

2

cos( = 54°)

=

2

cos( = 63°)

–2 – Φ 1

2–φ 2

2 –

2–φ

1

1

φ

2

2

2

cos( = 72°)

=

=

2 2–Φ

–2 + φ cos( = 81°)

1

2

2

2+Φ

–

This pattern uses the identities phi = φ = √2 – Φ and Phi = Φ = √2 + φ together with the half-angle formula for cos(A/2) (see below) starting from cos(36)=Phi/2 and cos(72)=phi/2. The pattern continues with the cosines of 4.5°, 13.5°, etc.

The √(2 ± √u) pattern Ernesto La Orden also put many angles into this pattern:

An gle 90

cosine sine

An gle

√2 – √4

= 0

2 75

0

An gle 72

cosine sine √2 – Phi

15

54

√2 – phi

2

60

= 2

45

= 2

30

36

30

18

√2 + √1

=

1

45

√3

60

2

22. √2 + √2 5

67. 5

2

√2 + √3

75 2

0

√2 + √4

= 1

2

√2 + phi

54 2

√2 + Phi

72 2

√2

2

15

1

22. 5

2

√2 – √0

36 2

2

√2 – √1

18 2

√2 – √3

67. √2 – √2 5

An gle

90

The table on the right has values of u that are Phi2 = 2.618033.. and phi2 = 0.381966..

Proofs 30° 45° and 60° Here are two simple triangles which give us the formulae for the trig values of these three angles:-

This triangle is just a square cut along a diagonal. If the sides are of length 1, the diagonal is length √2. This Here is an equilateral triangle where all sides and all angles are equal (to 60°). If the sides are of length 2, gives the sin, cos and tan of 45°. then when we cut it in half as shown, the two triangles have 60°, 30° and 90° angles with a side of length 1 and a hypotenuse of length 2. The other side is therefore of length √3. So we can read off the sin cos and tan of both 30° and 60°.

36° and 54°, 18° and 72° For 36° and 72° we need some further work based on the geometry of a regular pentagon which has angles of 36° and 72°. If the sides of the pentagon are of length 1, the diagonals are of the golden section number in length Phi where: 1 + √5

Phi = = 1.618033 988.. =

2

1

= 1 +

Phi

The upper triangle with angles 72°, 72° and 36° and sides of lengths 1, Phi and Phi shows the trig values for 18° and 72°. The lower triangle with angles of 36°, 36° and 108° and sides of lengths 1, 1 and Phi shows the trig values of 36° and 54°.

15° and 75° If we take the triangle on the left, we can calculate the length of the third side using the Cosine Formula. If, in a triangle with sides a, b and c, we know both sides b and c and also the angle A between sides b and c then we can compute the length of third side, a, as follows: a2 = b2 + c2 – 2 b c cos(A) For our triangle on the left, the known sides are b=2 and c=2 and the angle between them is A=30°. The length of the third side, the base a, is therefore: a2 = 22 + 22 – 2 x 2 x 2 x cos(30°) = 8 – 4 √3

= 2 (4 – 2 √3) But (√3 – 1) = 3 + 1 – 2 √3 = 4 – 2 √3 and so 2

a2 = 2 (√3 – 1)2 Taking the square-root: a = √2 (√3 – 1) which we can also write as = 2 (√3 – 1) / √2 Using this expressions for a, we can expand the triangle by a factor of √2, to get rid of the denominator. Finally, we put in a line from the top of the triangle to the centre of the base a to make two right-angled triangles. This will halve the side a and cut the triangle into two and gets rid of the factor 2 also. We then arrive at the triangle on the right which shows the sines and cosines of 75° and 15°: Ailles Rectangle

An alternative (easier) method for sine and cosine of 15° and 75° is found in Ailles Rectangle (named after an Ontario high school teacher). It is easy to because it is two (green) 45° right-angled triangles stuck onto the sides of a (white) 30-60-90 triangle and the rectangle completed with a (yellow) 15-75-90 triangle on the hypotenuse of the 30-60-90 triangle as shown here. The 30-60-90 sides are "as usual", namely 1, 2 and √3. From the two 45-45-90 triangles, it is quite easy to see that x is √3/√2 and y is 1/√2 from which we can read off the sines and cosines of 15° and 75°.

Trig Formulae Many symmetries and patterns are apparent in the table. They reflect some underlying identities such as: sin(x) = cos(90° – x) ta n( x) = tan sin(x) = a / h 2(x cos(x) = b / ) + 1 h tan(x) = a / b = cot(x) = b / a

sin2(x) + cos2(x) = 1 sin(x)

1

co t(x )=

cos(x) 1

cos(x) =

tan(x)

sin(x) 1

cot2(x )+1 =

sin2(x)

cos2(x)

If we know the value of a trig function on two angles A and B, we can determine the trig function values of their sum and difference using the following identities: sin( A + B ) = sin(A)cos(B) + cos(A)sin(B) sin( A – B ) = sin(A)cos(B) – cos(A)sin(B) cos( A + B ) = cos(A)cos(B) – sin(A)sin(B) cos( A – B ) = cos(A)cos(B) + sin(A)sin(B)

tan(A) + tan(B)

tan(A + B) =

1 – tan(A) tan(B) tan(A) – tan(B)

tan(A – B) =

1 + tan(A) tan(B)

If the two angles are the same (i.e. A=B) we get the sines and cosines of double the angle. Rearranging those formulae gives the formula for the sin or cosine of half an angle: sin( 2A ) = 2 sin(A) cos(A) cos( 2A ) = 1 – 2 sin2(A) cos( 2A ) = cos2(A) – sin2(A) cos( 2A ) = 2 cos2(A) – 1

A si n

2

1 – cos(A) = 2

A

c o s ta n ( 2 A )

2 tan(A)

1 + cos(A) =

2

2

A

sin(A)

1 – cos(A)

2

1 + cos(A)

sin(A)

1 – tan2(A)

=

t a n

=

=

A diagram to relate many angles and Phi Robert Gray's page on Coordinates for many regular solids has an amazing diagram at the bottom which relates Phi to the angles of 18°, 30°, 36°, 45°, 54°, 60° and 72° according to their 3D coordinates in the solids. Each of those angles is measured from the top most point of the circle when a vertical line is turned through that angle. Each line from the base point meets the circle at a point whose a height is 1 (72°), 1+Phi (60°), 2+Phi (54°), 2+2 Phi (45°), 2+3 Phi (36°), 3+3 Phi (30°) or 3+4 Phi (18°). Do look at his pages for more fascinating information on 120 3D solids, of which we will also explore the most symmetrical 5 on our next page.

Things to do 1. Suppose the origin of the circle is the lowest point and its radius is 2 + 2 Phi. Find the equation of the circle. 2. Use your answer to the previous question to find the coordinates of each of the points on the circle with the angles shown. 3. Compute the lengths of each of the red lines from the lowest point to the points shown on the circle. 4. From any two points A and B on a circle, the angle AOB at the centre of a circle, O, is twice the angle at any point on the circumference in the same sector. In the diagram,all the red angles at the circumference are equal; the red angles are twice the blue angle AOB at the centre; the red angles are to a point in the same sector of the circle as is the centre of the circle so they cannot be in the grey sector.

Use the above theorem to find three points on the circle ABOVE this Things To Do section where a line from the centre makes an angle with the vertical of i. 2×18=36° ii. 2×30=60° iii. 2×36=72°

Other angles with exact trig expressions involving square-roots Are there other angles with a simple exact expression for their cosine or sine? Well it all depends upon what you mean by simple! Carl Friedrich GAUSS (177 - 1855) looked at a similar problem which answers this question. He investigated if there was a method of constructing a regular polygon of n sides using only a pair of comes (to draw circles) and a straight-edge (a ruler with no markings). We know we can construct a regular polygon for all of the values of n=3, 4, 5, 6, 8 and 10. Halving There is a simple geometrical way to use comes to divide an angle into two (angle bisection). So all the angles in a regular n-gon can be split into two to make a regular 2n-gon. We can repeat the process to get a 4n-gon, 8n-gon and in general a 2kn-gon for any k once we have a method of constructing a regular n-gon. The Trig Formula section above contains a formula for the cosine of half an angle in of the cosine of the (whole) angle: c o s

A

1 + cos(A) =

2

2 + 2cos(A) =

2

2

As Mitch Wyatt pointed out to me, since we know that cos(90°) is 0 and 90° is /2 radians, we can use it to find the cosine of half that angle (45° or /4 radians) and then halve that angle again and so on. Each time we introduce another square root so we get a cascading or nested sequence of square roots: co s co

√ 2 4

= 2

s

√2√ 2 = +

8

2 co s

√ 2√2√ 2 + + =

16

2 co s

√

√ 2 2√ 2√ 2 = +++

32

2

However, this page is about sines and cosines which have simpler expressions, so we will not expand on this except to say that it shows how we can always find an exact expression for the sine (or cosine) of 1/2, 1/4, 1/8, ..., 1/2 , ... of any angle for which we have an exact sine (or cosine) expression. n

Superimposing If we construct a regular triangle (3 sides) and with the same circle centre, construct three regular pentagons (5 sides) with each having one vertex in common with the triangle, we will have the 15 vertices of a regular 15-gon. This is shown on the right with the 3 pentagons in blue on the same circle, each having a vertex in common with the red triangle and the regular 15-gon appears in yellow. By superimposing two regular polygons like this, we can construct a regular P×Q-gon (if P and Q have no factors in common otherwise more than one vertex of each will coincide). Do we know all the angles? All this was known in Euclid's time, around the year 300 BC. So what about 7ths and 9ths? Is it possible to find sines and cosines of all the multiples of 1/7 and 1/9 of a turn in exact (using square roots)? What about 11ths and 12ths etc.? In the next 2000 years no one found an exact geometric method for 7-gons or 9-gons but also no one had proved it was impossible to construct such regular polygons. Then C F Gauss completely solved the problem while he was a student at Göttingen between 1795 and 1798. Gauss found the conditions on n and its the prime factors to solve two equivalent problems: • drawing a regular n-sided polygon using only a straight edge and com and •

expressing the cos and sin of 360/n° using only square roots. If we factor n as 2ap1bp2c..., i.e. a, b, c, ... are the powers of n's prime factors: 2, p1, p2, ... (the prime's power is 0 if it is not a factor of n) then both of the problems are solvable when

•

b,c,... and all the powers except a, the power of 2, must be 1, and

•

the primes>2 that are factors of n (that is p1, p2, ...) must be of the form 22 +1 for some number k. k

Both problems are solvable for these values of n and only for these values. Prime numbers of the form 22 +1 are called Fermat primes. The series of numbers of the form 22 +1 begins k

k

22 + 1 = 3, 22 + 1 = 5, 22 + 1 = 17, 22 + 1 = 257, 22 + 1 = 65537, ... However not every number of the form 22 + 1 is prime -- and it is only the prime ones that we must have as factors of n. The next one, 22 + 1 is 4294967297 and has a factor of 641 so it is not prime. In fact, we do not know if there are any more primes of this form except the first 5 listed above. 0

1

2

3

4

k

5

Such numbers, n, of the form Gauss describes are as follows, one per row, each a product of some of the Fermat primes (but each prime at most once) followed by its multiples of two. For any number in the table, its double is also in the table: 24 36 5 10 3×5=15 30 17 34 3×17=51 102 5×17=75 150 3×5×17=225 450 257 514

8 12 20 60 68 204 300 900 1028

16 24 40 120 136 408 600 1800 2056

32 48 80 240 ... ... ... ... ...

... ... ... ...

When put in order, we have (1), 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, ... which is Sloane's A003401. Here, for instance, is the cosine of an angle involving 17ths:

8

co s

√1 √ 7 – 68 + 12√17 + 2(√17 – 1)S – 16 T 1 + =S –

17 wher√ e S 34 – 2√17 =

16 an √ d T 34 + 2√17 =

Things to do 1.

From the five 'starting values' above: 3, 5, 17, 257 and 65537 we can multiply these and also double any number to get a constructible polygon or an angle with a sine which involves nothing more than square-roots. Don't forget that we can double 2 to get 4 (a square), 8 (octagon), 16, etc too but we cannot use any of the 5 odd primes more than once in any product (so 3x3=9 and 5x5=25 are not in the list but 3x5x7 is). A complete list will involve these five numbers of course together with their products 3x5=15, 3x17=51, ... and you can double any of these any number of times: 2x3=6, 2x15=30 as well as 2x6=12, 2x12=24,... . a.

What are the first 12 values in the list that starts 3, 4, 5, ...?

b. Check that there are 24 values (excluding 1 and 2) less than 100. c.

Is 100 in the list?

2.

From the five known values: 3, 5, 17, 257 and 65537, there are a finite number of odd numbers n for which sin(360/n°) can be written with square-roots alone i.e. products of these 5 numbers where no number can be used more than once. How many odd numbers (excluding 1) can you make using these 5 no more than once in each product?

3.

This is the total number of known polygons we can construct with ruler and com or which have a sine (cosine) formed from nothing more than square-roots. With thanks to Richard Duffy for suggesting this puzzle

Tom Ace pointed out that there is more about this in chapter 15 of Oystein Ore's Number Theory and Its History from 1948 and now available as a Dover book(1988).

A Table of Exact Trig values

that are expressible as simple involving square-roots. a

b

rad deg ian ree s s

0

0

cos(a) sin(b)

tan(a) cot(b)

1

0

90

√6 – √3 + √2 – 2 (√2 – 1)(√3 – √2)

82. 5

2 – √3 tan(15°) = [0; 3, [1, 2]] tan2(15°) = [0; 13, [1, 12]]

75

5π 12

72

2π 5

π 7.5

24

√6

π 15

4

12

√2 +

1 4

=

√ 3 +

4 =

+

2 4 cos2(15°) = [0; 1, 13, [1, 12]] 1√ 05

π 18

10

5

√ 12

8

8

=

π 2 11

π 24

√ 5 +

+

deg rad ree ian s s

8

1 8

2

–

2√5 5 tan2(18°) = [0; 9, [2,8]]

4 cos2(18°) = [0; 1, 9, [2, 8]] 2 + √2

4

π 22· 8

5

2

=

√ 8 +

8 cos2(22·5°) = [0; 1, 5, [1, 4]]

8

√2 – 1 tan(22·5) = [0; [2]] tan2(22·5) = [0; 5, [1,4]]

67· 3π 5 8

3

π 30

=

6

4

√3

√3

2

3

cos(30°) = [0; 1, 6, [2, 6]] cos2(30°) = [0; 1, 3] √5

π 36

1 +

4

5

4

3

3

√5

=

+

8

8

√ 5 – 2√5 tan2(36°) = [0; 1, 1, [8, 2]]

5π 37. 24 5

√6 + √3 – √2 – 2 (√2 + 1)(√3 – √2)

54

3π 10

52. 7π 5 24

1

2 =

4

π

tan(30°) = [0; 1, [1,2]] tan2(30°) = [0; 3]

cos(36°) = [0; 1, [4]] cos2(36°) = [0; 1, 1, 1, [8, 2]]

π 45

60

√2

4

1

45

√6 – √3 – √2 + 2 (√2 – 1)(√3 + √2)

52. 5

π 2

cos(45°) = [0; 1, [2]] cos2(45°) = [0; 2] 7π 52. 24 5 1√ 05 3π 54 10

5

=

π 24

√ 5 –

–

5

1

8

8

2

+

2√5 5

36

π

30

π

5

tan2(54°) = [1; 1, [8, 2]] 4 cos2(54°) = [0; 2, 1, [8, 2]] 1

1

π 60

=

3

2

4 cos(60°) = [0; 2] cos2(60°) = [0; 4]

3π 67· 2 – √2 8 5

4 = 2

8 –

√3 tan(60°) = [1; [1,2]] tan2(60°) = 3

1 + √2 tan(67·5) = [2; [2]] tan2(67·5) = [5; [1,4]]

22· 5

6

π 8

8 cos (67·5) = [0; 6, [1, 4]]

8

2

√5 2π 72 5

1 –

4

3

√5

=

4

–

8

√ 5 + 2√5

8

tan2(72°) = [9; [2, 8]]

cos(72°) = [0; 3, [4]] cos2(72°) = [0; 10, [2, 8] ] √6 5π 75 12

4

√2 –

1 =

4

√ 3 –

4 =

2 4 2 cos (75°) = [0; 14, [1, 12]]

√ 12 –

8

8

11

π 82. 5

24

π 90

0

2

18

2 + √3 tan(75°) = [3; [1, 2]] tan2(75°) = [13; [1,12]]

15

√6 + √3 + √2 + 2 (√2 + 1)(√3 + √2)

7.5

Infinity

0

π 10

π 12

π 24 0

The values in the table are those angles of the form n° or n/2 or n/3 for a whole number n, between 0 and 90° whose sin or cosine is rational, or whose continued fraction is periodic or the square of the trig value has a periodic continued fraction. 1

Continued fraction [a; b,c,d,...] means a + 1 b+ 1 c+ d + ...

and the periodic continued fraction [a;b, c, d, e, d, e, d, e, d, e,...] is written as [a; b, c, [d, e]].

Trig functions of Angles <0 or >90° Top of Form

To find the trig. values of all angles including those bigger than 90 degrees and negative angles: 1. select a trig function 2. type the angle in the box and then 3.

click on the button

to find which angle in the range 0-90° has the same value:

Select a trig function:

(

°)

Degrees-Radians Converter To convert between DEGREES and RADIANS: 1. enter the angle as a number in one box leaving the other empty 2.

then click the button to do the conversion

You can use Pi in the radians box and * for multiplication e.g. 3*Pi/2: degrees

radians

Bottom of Form

Patterns The Simple Square-Root pattern Ernesto La Orden of Madrid pointed out the following neat way to connect and the easiest of the sines (cosines): Angl e

sine cosine √4

90

30

= 1

2

30

2 √2

1

=

2

√2

√1

1

=

2

2 √0

0

=0

2

The √(2 ± Phi) pattern cos( = 9°)

0

√3

60

45

Angl e

1

2

2 +

2+Φ

45

60

90

1

1

2

2

cos( = 18°)

=

2

cos( = 27°)

+2 + φ 1

2+Φ 2

2 +

2–φ

1

1

Φ

2

2

2

cos( = 36°)

=

=

2 2+φ

+2 – Φ cos( = 54°)

1 2

2 2–Φ –

1 =

2

2–φ

1 2

cos( = 63°)

2

cos( = 72°)

1 2

2 2+φ –

– 2–φ 1 =

2

2–Φ

φ =

2

1 2

cos( = 81°)

2 – 2+Φ

This pattern uses the identities phi = φ = √2 – Φ and Phi = Φ = √2 + φ together with the half-angle formula for cos(A/2) (see below) starting from cos(36)=Phi/2 and cos(72)=phi/2. The pattern continues with the cosines of 4.5°, 13.5°, etc.

The √(2 ± √u) pattern Ernesto La Orden also put many angles into this pattern:

An gle 90

cosine sine

An gle

√2 – √4

= 0

2 75

0

An gle 72

cosine sine √2 – Phi

15

54

√2 – phi

2

60

= 2

45

= 2

30

36

30

18

√2 + √1

=

1

45

√3

60

2

22. √2 + √2 5

67. 5

2

√2 + √3

75 2

0

√2 + √4

= 1

2

√2 + phi

54 2

√2 + Phi

72 2

√2

2

15

1

22. 5

2

√2 – √0

36 2

2

√2 – √1

18 2

√2 – √3

67. √2 – √2 5

An gle

90

The table on the right has values of u that are Phi2 = 2.618033.. and phi2 = 0.381966..

Proofs 30° 45° and 60° Here are two simple triangles which give us the formulae for the trig values of these three angles:-

This triangle is just a square cut along a diagonal. If the sides are of length 1, the diagonal is length √2. This Here is an equilateral triangle where all sides and all angles are equal (to 60°). If the sides are of length 2, gives the sin, cos and tan of 45°. then when we cut it in half as shown, the two triangles have 60°, 30° and 90° angles with a side of length 1 and a hypotenuse of length 2. The other side is therefore of length √3. So we can read off the sin cos and tan of both 30° and 60°.

36° and 54°, 18° and 72° For 36° and 72° we need some further work based on the geometry of a regular pentagon which has angles of 36° and 72°. If the sides of the pentagon are of length 1, the diagonals are of the golden section number in length Phi where: 1 + √5

Phi = = 1.618033 988.. =

2

1

= 1 +

Phi

The upper triangle with angles 72°, 72° and 36° and sides of lengths 1, Phi and Phi shows the trig values for 18° and 72°. The lower triangle with angles of 36°, 36° and 108° and sides of lengths 1, 1 and Phi shows the trig values of 36° and 54°.

15° and 75° If we take the triangle on the left, we can calculate the length of the third side using the Cosine Formula. If, in a triangle with sides a, b and c, we know both sides b and c and also the angle A between sides b and c then we can compute the length of third side, a, as follows: a2 = b2 + c2 – 2 b c cos(A) For our triangle on the left, the known sides are b=2 and c=2 and the angle between them is A=30°. The length of the third side, the base a, is therefore: a2 = 22 + 22 – 2 x 2 x 2 x cos(30°) = 8 – 4 √3

= 2 (4 – 2 √3) But (√3 – 1) = 3 + 1 – 2 √3 = 4 – 2 √3 and so 2

a2 = 2 (√3 – 1)2 Taking the square-root: a = √2 (√3 – 1) which we can also write as = 2 (√3 – 1) / √2 Using this expressions for a, we can expand the triangle by a factor of √2, to get rid of the denominator. Finally, we put in a line from the top of the triangle to the centre of the base a to make two right-angled triangles. This will halve the side a and cut the triangle into two and gets rid of the factor 2 also. We then arrive at the triangle on the right which shows the sines and cosines of 75° and 15°: Ailles Rectangle

An alternative (easier) method for sine and cosine of 15° and 75° is found in Ailles Rectangle (named after an Ontario high school teacher). It is easy to because it is two (green) 45° right-angled triangles stuck onto the sides of a (white) 30-60-90 triangle and the rectangle completed with a (yellow) 15-75-90 triangle on the hypotenuse of the 30-60-90 triangle as shown here. The 30-60-90 sides are "as usual", namely 1, 2 and √3. From the two 45-45-90 triangles, it is quite easy to see that x is √3/√2 and y is 1/√2 from which we can read off the sines and cosines of 15° and 75°.

Trig Formulae Many symmetries and patterns are apparent in the table. They reflect some underlying identities such as: sin(x) = cos(90° – x) ta n( x) = tan sin(x) = a / h 2(x cos(x) = b / ) + 1 h tan(x) = a / b = cot(x) = b / a

sin2(x) + cos2(x) = 1 sin(x)

1

co t(x )=

cos(x) 1

cos(x) =

tan(x)

sin(x) 1

cot2(x )+1 =

sin2(x)

cos2(x)

If we know the value of a trig function on two angles A and B, we can determine the trig function values of their sum and difference using the following identities: sin( A + B ) = sin(A)cos(B) + cos(A)sin(B) sin( A – B ) = sin(A)cos(B) – cos(A)sin(B) cos( A + B ) = cos(A)cos(B) – sin(A)sin(B) cos( A – B ) = cos(A)cos(B) + sin(A)sin(B)

tan(A) + tan(B)

tan(A + B) =

1 – tan(A) tan(B) tan(A) – tan(B)

tan(A – B) =

1 + tan(A) tan(B)

If the two angles are the same (i.e. A=B) we get the sines and cosines of double the angle. Rearranging those formulae gives the formula for the sin or cosine of half an angle: sin( 2A ) = 2 sin(A) cos(A) cos( 2A ) = 1 – 2 sin2(A) cos( 2A ) = cos2(A) – sin2(A) cos( 2A ) = 2 cos2(A) – 1

A si n

2

1 – cos(A) = 2

A

c o s ta n ( 2 A )

2 tan(A)

1 + cos(A) =

2

2

A

sin(A)

1 – cos(A)

2

1 + cos(A)

sin(A)

1 – tan2(A)

=

t a n

=

=

A diagram to relate many angles and Phi Robert Gray's page on Coordinates for many regular solids has an amazing diagram at the bottom which relates Phi to the angles of 18°, 30°, 36°, 45°, 54°, 60° and 72° according to their 3D coordinates in the solids. Each of those angles is measured from the top most point of the circle when a vertical line is turned through that angle. Each line from the base point meets the circle at a point whose a height is 1 (72°), 1+Phi (60°), 2+Phi (54°), 2+2 Phi (45°), 2+3 Phi (36°), 3+3 Phi (30°) or 3+4 Phi (18°). Do look at his pages for more fascinating information on 120 3D solids, of which we will also explore the most symmetrical 5 on our next page.

Things to do 1. Suppose the origin of the circle is the lowest point and its radius is 2 + 2 Phi. Find the equation of the circle. 2. Use your answer to the previous question to find the coordinates of each of the points on the circle with the angles shown. 3. Compute the lengths of each of the red lines from the lowest point to the points shown on the circle. 4. From any two points A and B on a circle, the angle AOB at the centre of a circle, O, is twice the angle at any point on the circumference in the same sector. In the diagram,all the red angles at the circumference are equal; the red angles are twice the blue angle AOB at the centre; the red angles are to a point in the same sector of the circle as is the centre of the circle so they cannot be in the grey sector.

Use the above theorem to find three points on the circle ABOVE this Things To Do section where a line from the centre makes an angle with the vertical of i. 2×18=36° ii. 2×30=60° iii. 2×36=72°

Other angles with exact trig expressions involving square-roots Are there other angles with a simple exact expression for their cosine or sine? Well it all depends upon what you mean by simple! Carl Friedrich GAUSS (177 - 1855) looked at a similar problem which answers this question. He investigated if there was a method of constructing a regular polygon of n sides using only a pair of comes (to draw circles) and a straight-edge (a ruler with no markings). We know we can construct a regular polygon for all of the values of n=3, 4, 5, 6, 8 and 10. Halving There is a simple geometrical way to use comes to divide an angle into two (angle bisection). So all the angles in a regular n-gon can be split into two to make a regular 2n-gon. We can repeat the process to get a 4n-gon, 8n-gon and in general a 2kn-gon for any k once we have a method of constructing a regular n-gon. The Trig Formula section above contains a formula for the cosine of half an angle in of the cosine of the (whole) angle: c o s

A

1 + cos(A) =

2

2 + 2cos(A) =

2

2

As Mitch Wyatt pointed out to me, since we know that cos(90°) is 0 and 90° is /2 radians, we can use it to find the cosine of half that angle (45° or /4 radians) and then halve that angle again and so on. Each time we introduce another square root so we get a cascading or nested sequence of square roots: co s co

√ 2 4

= 2

s

√2√ 2 = +

8

2 co s

√ 2√2√ 2 + + =

16

2 co s

√

√ 2 2√ 2√ 2 = +++

32

2

However, this page is about sines and cosines which have simpler expressions, so we will not expand on this except to say that it shows how we can always find an exact expression for the sine (or cosine) of 1/2, 1/4, 1/8, ..., 1/2 , ... of any angle for which we have an exact sine (or cosine) expression. n

Superimposing If we construct a regular triangle (3 sides) and with the same circle centre, construct three regular pentagons (5 sides) with each having one vertex in common with the triangle, we will have the 15 vertices of a regular 15-gon. This is shown on the right with the 3 pentagons in blue on the same circle, each having a vertex in common with the red triangle and the regular 15-gon appears in yellow. By superimposing two regular polygons like this, we can construct a regular P×Q-gon (if P and Q have no factors in common otherwise more than one vertex of each will coincide). Do we know all the angles? All this was known in Euclid's time, around the year 300 BC. So what about 7ths and 9ths? Is it possible to find sines and cosines of all the multiples of 1/7 and 1/9 of a turn in exact (using square roots)? What about 11ths and 12ths etc.? In the next 2000 years no one found an exact geometric method for 7-gons or 9-gons but also no one had proved it was impossible to construct such regular polygons. Then C F Gauss completely solved the problem while he was a student at Göttingen between 1795 and 1798. Gauss found the conditions on n and its the prime factors to solve two equivalent problems: • drawing a regular n-sided polygon using only a straight edge and com and •

expressing the cos and sin of 360/n° using only square roots. If we factor n as 2ap1bp2c..., i.e. a, b, c, ... are the powers of n's prime factors: 2, p1, p2, ... (the prime's power is 0 if it is not a factor of n) then both of the problems are solvable when

•

b,c,... and all the powers except a, the power of 2, must be 1, and

•

the primes>2 that are factors of n (that is p1, p2, ...) must be of the form 22 +1 for some number k. k

Both problems are solvable for these values of n and only for these values. Prime numbers of the form 22 +1 are called Fermat primes. The series of numbers of the form 22 +1 begins k

k

22 + 1 = 3, 22 + 1 = 5, 22 + 1 = 17, 22 + 1 = 257, 22 + 1 = 65537, ... However not every number of the form 22 + 1 is prime -- and it is only the prime ones that we must have as factors of n. The next one, 22 + 1 is 4294967297 and has a factor of 641 so it is not prime. In fact, we do not know if there are any more primes of this form except the first 5 listed above. 0

1

2

3

4

k

5

Such numbers, n, of the form Gauss describes are as follows, one per row, each a product of some of the Fermat primes (but each prime at most once) followed by its multiples of two. For any number in the table, its double is also in the table: 24 36 5 10 3×5=15 30 17 34 3×17=51 102 5×17=75 150 3×5×17=225 450 257 514

8 12 20 60 68 204 300 900 1028

16 24 40 120 136 408 600 1800 2056

32 48 80 240 ... ... ... ... ...

... ... ... ...

When put in order, we have (1), 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, ... which is Sloane's A003401. Here, for instance, is the cosine of an angle involving 17ths:

8

co s

√1 √ 7 – 68 + 12√17 + 2(√17 – 1)S – 16 T 1 + =S –

17 wher√ e S 34 – 2√17 =

16 an √ d T 34 + 2√17 =

Things to do 1.

From the five 'starting values' above: 3, 5, 17, 257 and 65537 we can multiply these and also double any number to get a constructible polygon or an angle with a sine which involves nothing more than square-roots. Don't forget that we can double 2 to get 4 (a square), 8 (octagon), 16, etc too but we cannot use any of the 5 odd primes more than once in any product (so 3x3=9 and 5x5=25 are not in the list but 3x5x7 is). A complete list will involve these five numbers of course together with their products 3x5=15, 3x17=51, ... and you can double any of these any number of times: 2x3=6, 2x15=30 as well as 2x6=12, 2x12=24,... . a.

What are the first 12 values in the list that starts 3, 4, 5, ...?

b. Check that there are 24 values (excluding 1 and 2) less than 100. c. 2.

Is 100 in the list?

From the five known values: 3, 5, 17, 257 and 65537, there are a finite number of odd numbers n for which sin(360/n°) can be written with square-roots alone i.e. products of these 5 numbers where no number can be used more than once. How many odd numbers (excluding 1) can you make using these 5 no more than once in each product? This is the total number of known polygons we can construct with ruler and com or which have a sine (cosine) formed from nothing more than square-roots. With thanks to Richard Duffy for suggesting this puzzle

A Table of Exact Trig values that are expressible as simple involving square-roots. a

b

rad deg ian ree s s

0

0

cos(a) sin(b)

tan(a) cot(b)

1

0

90

√6 – √3 + √2 – 2 (√2 – 1)(√3 – √2)

82. 5

2 – √3 tan(15°) = [0; 3, [1, 2]] tan2(15°) = [0; 13, [1, 12]]

75

5π 12

72

2π 5

π 7.5

24

√6

π 15

4

12

√2 +

√ 3

1 4

=

+

4 =

+

2 4 cos2(15°) = [0; 1, 13, [1, 12]] 1√ 05

5

8

8

1

8

=

π 2 11

π 24

√ 5 +

π 18 +

10

√ 12

deg rad ree ian s s

8

2

–

2√5 5 tan2(18°) = [0; 9, [2,8]]

4 cos2(18°) = [0; 1, 9, [2, 8]] 2 + √2

4

π 22· 8

5

=

2

√ 8 +

8 cos (22·5°) = [0; 1, 5, [1, 4]]

8

√2 – 1 tan(22·5) = [0; [2]] tan2(22·5) = [0; 5, [1,4]]

67· 3π 5 8

2

π 30

3

√3

√3

60

π

=

4

6

3

2

cos(30°) = [0; 1, 6, [2, 6]] cos2(30°) = [0; 1, 3] √5

π 36

1 +

4

5

4

tan(30°) = [0; 1, [1,2]] tan2(30°) = [0; 3] 3

√5

=

+

8

8

√ 5 – 2√5 tan2(36°) = [0; 1, 1, [8, 2]]

cos(36°) = [0; 1, [4]] 2 cos (36°) = [0; 1, 1, 1, [8, 2]] 5π 37. 24 5

√6 + √3 – √2 – 2 (√2 + 1)(√3 – √2)

=

4

54

3π 10

52. 7π 5 24

1

2

π 45

3

√2

4

1

45

√6 – √3 – √2 + 2 (√2 – 1)(√3 + √2)

52. 5

π 2

cos(45°) = [0; 1, [2]] cos2(45°) = [0; 2] 7π 52. 24 5 1√ 05 3π 54 10

– =

π 24

√ 5

5

–

5

1

8

8

2

+

2√5 5

36

π

30

π

5

tan2(54°) = [1; 1, [8, 2]] 4 cos2(54°) = [0; 2, 1, [8, 2]] 1

1

π 60

=

3

√3 tan(60°) = [1; [1,2]] tan2(60°) = 3

2

4 cos(60°) = [0; 2] cos2(60°) = [0; 4]

3π 67· 2 – √2 8 5

4 2

=

1 + √2 tan(67·5) = [2; [2]] tan2(67·5) = [5; [1,4]]

√ 8 –

8

8

22· 5

6

π 8

cos2(67·5) = [0; 6, [1, 4]] √5 2π 72 5

1 –

4

3

√5

=

4

–

8

√ 5 + 2√5

8

tan2(72°) = [9; [2, 8]]

cos(72°) = [0; 3, [4]] cos2(72°) = [0; 10, [2, 8] ] √6 5π 75 12

4

√2 –

1 =

4

√ 3 –

4 =

2 4 cos2(75°) = [0; 14, [1, 12]]

√ 12 –

8

8

11

π 82. 5

24

π 90

0

2

18

2 + √3 tan(75°) = [3; [1, 2]] tan2(75°) = [13; [1,12]]

15

√6 + √3 + √2 + 2 (√2 + 1)(√3 + √2)

7.5

Infinity

0

π 10

π 12

π 24 0

The values in the table are those angles of the form n° or n/2 or n/3 for a whole number n, between 0 and 90° whose sin or cosine is rational, or whose continued fraction is periodic or the square of the trig value has a periodic continued fraction. 1

Continued fraction [a; b,c,d,...] means a + 1 b+ 1 c+ d + ...

and the periodic continued fraction [a;b, c, d, e, d, e, d, e, d, e,...] is written as [a; b, c, [d, e]].

Trig functions of Angles <0 or >90° Top of Form

To find the trig. values of all angles including those bigger than 90 degrees and negative angles: 1. select a trig function 2. type the angle in the box and then 3.

click on the button

to find which angle in the range 0-90° has the same value:

Select a trig function:

(

°)

Degrees-Radians Converter To convert between DEGREES and RADIANS: 1. enter the angle as a number in one box leaving the other empty 2.

then click the button to do the conversion

You can use Pi in the radians box and * for multiplication e.g. 3*Pi/2: degrees

radians

Bottom of Form

Patterns The Simple Square-Root pattern Ernesto La Orden of Madrid pointed out the following neat way to connect and the easiest of the sines (cosines): Angl e

sine cosine √4

90

30

= 1

2

30

2 √2

1

=

2

√2

√1

1

=

2

2 √0

0

=0

2

The √(2 ± Phi) pattern cos( = 9°)

0

√3

60

45

Angl e

1

2

2 +

2+Φ

45

60

90

1

1

2

2

cos( = 18°)

=

2

cos( = 27°)

+2 + φ 1

2+Φ 2

2 +

2–φ

1

1

Φ

2

2

2

cos( = 36°)

=

=

2 2+φ

+2 – Φ cos( = 54°)

1 2

2 2–Φ –

1 =

2

2–φ

1 2

cos( = 63°)

2

cos( = 72°)

1 2

2 2+φ –

– 2–φ 1 =

2

2–Φ

φ =

2

1 2

cos( = 81°)

2 – 2+Φ

This pattern uses the identities phi = φ = √2 – Φ and Phi = Φ = √2 + φ together with the half-angle formula for cos(A/2) (see below) starting from cos(36)=Phi/2 and cos(72)=phi/2. The pattern continues with the cosines of 4.5°, 13.5°, etc.

The √(2 ± √u) pattern Ernesto La Orden also put many angles into this pattern:

An gle 90

cosine sine

An gle

√2 – √4

= 0

2 75

0

An gle 72

cosine sine √2 – Phi

15

54

√2 – phi

2

60

= 2

45

= 2

30

36

30

18

√2 + √1

=

1

45

√3

60

2

22. √2 + √2 5

67. 5

2

√2 + √3

75 2

0

√2 + √4

= 1

2

√2 + phi

54 2

√2 + Phi

72 2

√2

2

15

1

22. 5

2

√2 – √0

36 2

2

√2 – √1

18 2

√2 – √3

67. √2 – √2 5

An gle

90

The table on the right has values of u that are Phi2 = 2.618033.. and phi2 = 0.381966..

Proofs 30° 45° and 60° Here are two simple triangles which give us the formulae for the trig values of these three angles:-

This triangle is just a square cut along a diagonal. If the sides are of length 1, the diagonal is length √2. This Here is an equilateral triangle where all sides and all angles are equal (to 60°). If the sides are of length 2, gives the sin, cos and tan of 45°. then when we cut it in half as shown, the two triangles have 60°, 30° and 90° angles with a side of length 1 and a hypotenuse of length 2. The other side is therefore of length √3. So we can read off the sin cos and tan of both 30° and 60°.

36° and 54°, 18° and 72° For 36° and 72° we need some further work based on the geometry of a regular pentagon which has angles of 36° and 72°. If the sides of the pentagon are of length 1, the diagonals are of the golden section number in length Phi where: 1 + √5

Phi = = 1.618033 988.. =

2

1

= 1 +

Phi

The upper triangle with angles 72°, 72° and 36° and sides of lengths 1, Phi and Phi shows the trig values for 18° and 72°. The lower triangle with angles of 36°, 36° and 108° and sides of lengths 1, 1 and Phi shows the trig values of 36° and 54°.

15° and 75° If we take the triangle on the left, we can calculate the length of the third side using the Cosine Formula. If, in a triangle with sides a, b and c, we know both sides b and c and also the angle A between sides b and c then we can compute the length of third side, a, as follows: a2 = b2 + c2 – 2 b c cos(A) For our triangle on the left, the known sides are b=2 and c=2 and the angle between them is A=30°. The length of the third side, the base a, is therefore: a2 = 22 + 22 – 2 x 2 x 2 x cos(30°) = 8 – 4 √3

= 2 (4 – 2 √3) But (√3 – 1) = 3 + 1 – 2 √3 = 4 – 2 √3 and so 2

a2 = 2 (√3 – 1)2 Taking the square-root: a = √2 (√3 – 1) which we can also write as = 2 (√3 – 1) / √2 Using this expressions for a, we can expand the triangle by a factor of √2, to get rid of the denominator. Finally, we put in a line from the top of the triangle to the centre of the base a to make two right-angled triangles. This will halve the side a and cut the triangle into two and gets rid of the factor 2 also. We then arrive at the triangle on the right which shows the sines and cosines of 75° and 15°: Ailles Rectangle

An alternative (easier) method for sine and cosine of 15° and 75° is found in Ailles Rectangle (named after an Ontario high school teacher). It is easy to because it is two (green) 45° right-angled triangles stuck onto the sides of a (white) 30-60-90 triangle and the rectangle completed with a (yellow) 15-75-90 triangle on the hypotenuse of the 30-60-90 triangle as shown here. The 30-60-90 sides are "as usual", namely 1, 2 and √3. From the two 45-45-90 triangles, it is quite easy to see that x is √3/√2 and y is 1/√2 from which we can read off the sines and cosines of 15° and 75°.

Trig Formulae Many symmetries and patterns are apparent in the table. They reflect some underlying identities such as: sin(x) = cos(90° – x) ta n( x) = tan sin(x) = a / h 2(x cos(x) = b / ) + 1 h tan(x) = a / b = cot(x) = b / a

sin2(x) + cos2(x) = 1 sin(x)

1

co t(x )=

cos(x) 1

cos(x) =

tan(x)

sin(x) 1

cot2(x )+1 =

sin2(x)

cos2(x)

If we know the value of a trig function on two angles A and B, we can determine the trig function values of their sum and difference using the following identities: sin( A + B ) = sin(A)cos(B) + cos(A)sin(B) sin( A – B ) = sin(A)cos(B) – cos(A)sin(B) cos( A + B ) = cos(A)cos(B) – sin(A)sin(B) cos( A – B ) = cos(A)cos(B) + sin(A)sin(B)

tan(A) + tan(B)

tan(A + B) =

1 – tan(A) tan(B) tan(A) – tan(B)

tan(A – B) =

1 + tan(A) tan(B)

If the two angles are the same (i.e. A=B) we get the sines and cosines of double the angle. Rearranging those formulae gives the formula for the sin or cosine of half an angle: sin( 2A ) = 2 sin(A) cos(A) cos( 2A ) = 1 – 2 sin2(A) cos( 2A ) = cos2(A) – sin2(A) cos( 2A ) = 2 cos2(A) – 1

A si n

2

1 – cos(A) = 2

A

c o s ta n ( 2 A )

2 tan(A)

1 + cos(A) =

2

2

A

sin(A)

1 – cos(A)

2

1 + cos(A)

sin(A)

1 – tan2(A)

=

t a n

=

=

A diagram to relate many angles and Phi Robert Gray's page on Coordinates for many regular solids has an amazing diagram at the bottom which relates Phi to the angles of 18°, 30°, 36°, 45°, 54°, 60° and 72° according to their 3D coordinates in the solids. Each of those angles is measured from the top most point of the circle when a vertical line is turned through that angle. Each line from the base point meets the circle at a point whose a height is 1 (72°), 1+Phi (60°), 2+Phi (54°), 2+2 Phi (45°), 2+3 Phi (36°), 3+3 Phi (30°) or 3+4 Phi (18°). Do look at his pages for more fascinating information on 120 3D solids, of which we will also explore the most symmetrical 5 on our next page.

Things to do 1. Suppose the origin of the circle is the lowest point and its radius is 2 + 2 Phi. Find the equation of the circle. 2. Use your answer to the previous question to find the coordinates of each of the points on the circle with the angles shown. 3. Compute the lengths of each of the red lines from the lowest point to the points shown on the circle. 4. From any two points A and B on a circle, the angle AOB at the centre of a circle, O, is twice the angle at any point on the circumference in the same sector. In the diagram,all the red angles at the circumference are equal; the red angles are twice the blue angle AOB at the centre; the red angles are to a point in the same sector of the circle as is the centre of the circle so they cannot be in the grey sector.

Use the above theorem to find three points on the circle ABOVE this Things To Do section where a line from the centre makes an angle with the vertical of i. 2×18=36° ii. 2×30=60° iii. 2×36=72°

Other angles with exact trig expressions involving square-roots Are there other angles with a simple exact expression for their cosine or sine? Well it all depends upon what you mean by simple! Carl Friedrich GAUSS (177 - 1855) looked at a similar problem which answers this question. He investigated if there was a method of constructing a regular polygon of n sides using only a pair of comes (to draw circles) and a straight-edge (a ruler with no markings). We know we can construct a regular polygon for all of the values of n=3, 4, 5, 6, 8 and 10. Halving There is a simple geometrical way to use comes to divide an angle into two (angle bisection). So all the angles in a regular n-gon can be split into two to make a regular 2n-gon. We can repeat the process to get a 4n-gon, 8n-gon and in general a 2kn-gon for any k once we have a method of constructing a regular n-gon. The Trig Formula section above contains a formula for the cosine of half an angle in of the cosine of the (whole) angle: c o s

A

1 + cos(A) =

2

2 + 2cos(A) =

2

2

As Mitch Wyatt pointed out to me, since we know that cos(90°) is 0 and 90° is /2 radians, we can use it to find the cosine of half that angle (45° or /4 radians) and then halve that angle again and so on. Each time we introduce another square root so we get a cascading or nested sequence of square roots: co s co

√ 2 4

= 2

s

√2√ 2 = +

8

2 co s

√ 2√2√ 2 + + =

16

2 co s

√

√ 2 2√ 2√ 2 = +++

32

2

However, this page is about sines and cosines which have simpler expressions, so we will not expand on this except to say that it shows how we can always find an exact expression for the sine (or cosine) of 1/2, 1/4, 1/8, ..., 1/2 , ... of any angle for which we have an exact sine (or cosine) expression. n

Superimposing If we construct a regular triangle (3 sides) and with the same circle centre, construct three regular pentagons (5 sides) with each having one vertex in common with the triangle, we will have the 15 vertices of a regular 15-gon. This is shown on the right with the 3 pentagons in blue on the same circle, each having a vertex in common with the red triangle and the regular 15-gon appears in yellow. By superimposing two regular polygons like this, we can construct a regular P×Q-gon (if P and Q have no factors in common otherwise more than one vertex of each will coincide). Do we know all the angles? All this was known in Euclid's time, around the year 300 BC. So what about 7ths and 9ths? Is it possible to find sines and cosines of all the multiples of 1/7 and 1/9 of a turn in exact (using square roots)? What about 11ths and 12ths etc.? In the next 2000 years no one found an exact geometric method for 7-gons or 9-gons but also no one had proved it was impossible to construct such regular polygons. Then C F Gauss completely solved the problem while he was a student at Göttingen between 1795 and 1798. Gauss found the conditions on n and its the prime factors to solve two equivalent problems: • drawing a regular n-sided polygon using only a straight edge and com and •

expressing the cos and sin of 360/n° using only square roots. If we factor n as 2ap1bp2c..., i.e. a, b, c, ... are the powers of n's prime factors: 2, p1, p2, ... (the prime's power is 0 if it is not a factor of n) then both of the problems are solvable when

•

b,c,... and all the powers except a, the power of 2, must be 1, and

•

the primes>2 that are factors of n (that is p1, p2, ...) must be of the form 22 +1 for some number k. k

Both problems are solvable for these values of n and only for these values. Prime numbers of the form 22 +1 are called Fermat primes. The series of numbers of the form 22 +1 begins k

k

22 + 1 = 3, 22 + 1 = 5, 22 + 1 = 17, 22 + 1 = 257, 22 + 1 = 65537, ... However not every number of the form 22 + 1 is prime -- and it is only the prime ones that we must have as factors of n. The next one, 22 + 1 is 4294967297 and has a factor of 641 so it is not prime. In fact, we do not know if there are any more primes of this form except the first 5 listed above. 0

1

2

3

4

k

5

Such numbers, n, of the form Gauss describes are as follows, one per row, each a product of some of the Fermat primes (but each prime at most once) followed by its multiples of two. For any number in the table, its double is also in the table: 24 36 5 10 3×5=15 30 17 34 3×17=51 102 5×17=75 150 3×5×17=225 450 257 514

8 12 20 60 68 204 300 900 1028

16 24 40 120 136 408 600 1800 2056

32 48 80 240 ... ... ... ... ...

... ... ... ...

When put in order, we have (1), 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, ... which is Sloane's A003401. Here, for instance, is the cosine of an angle involving 17ths:

8

co s

√1 √ 7 – 68 + 12√17 + 2(√17 – 1)S – 16 T 1 + =S –

17 wher√ e S 34 – 2√17 =

16 an √ d T 34 + 2√17 =

Things to do 1.

From the five 'starting values' above: 3, 5, 17, 257 and 65537 we can multiply these and also double any number to get a constructible polygon or an angle with a sine which involves nothing more than square-roots. Don't forget that we can double 2 to get 4 (a square), 8 (octagon), 16, etc too but we cannot use any of the 5 odd primes more than once in any product (so 3x3=9 and 5x5=25 are not in the list but 3x5x7 is). A complete list will involve these five numbers of course together with their products 3x5=15, 3x17=51, ... and you can double any of these any number of times: 2x3=6, 2x15=30 as well as 2x6=12, 2x12=24,... . a.

What are the first 12 values in the list that starts 3, 4, 5, ...?

b. Check that there are 24 values (excluding 1 and 2) less than 100. c. 2.

Is 100 in the list?

From the five known values: 3, 5, 17, 257 and 65537, there are a finite number of odd numbers n for which sin(360/n°) can be written with square-roots alone i.e. products of these 5 numbers where no number can be used more than once. How many odd numbers (excluding 1) can you make using these 5 no more than once in each product? This is the total number of known polygons we can construct with ruler and com or which have a sine (cosine) formed from nothing more than square-roots. With thanks to Richard Duffy for suggesting this puzzle

What angles have an exact expression for their sines, cosines and tangents? You might know that cos(60°)=1/2 and sin(60°)=√3/2 as well as tan(45°)=1, but are 30, 45 and 60 the only angles up to 90° with a formula for their trig values? No! There are lots more but not all angles have exact epxressions. Which angles do? What patterns are there in these expressions? This page shows expressions for many angles and even solves the complete problem of which angles do and which don't have exact trig expressions.

Contents of this Page The

icon means there is a Things to do investigation at the end of the section.

The sections marked with •

have an online interactive calculator.

A Table of Exact Trig values 

Converter for trig functions of angles <0 or >90°

•

Degress <-> Radians Converter

•

Patterns

•



The Simple Square-Root pattern



The √(2 ± Phi) pattern



The √(2 ± √u) pattern

Proofs 

30° 45° and 60°



36° and 54°, 18° and 72°



15° and 75° ○

Ailles Rectangle

•

Trig Formulae

•

A diagram to relate many angles and Phi

•

Other angles with exact trig expressions involving square-roots 

Halving



Superimposing



Do we know all the angles?

A Table of Exact Trig values that are expressible as simple involving square-roots. a

b

rad deg ian ree s s

0

0

cos(a) sin(b)

tan(a) cot(b)

1

0

90

√6 – √3 + √2 – 2 (√2 – 1)(√3 – √2)

82. 5

2 – √3 tan(15°) = [0; 3, [1, 2]] tan2(15°) = [0; 13, [1, 12]]

75

5π 12

72

2π 5

π 7.5

24

√6

π 15

√2 +

4

12

1 4

=

√ 3

4 =

+

+

2 4 cos2(15°) = [0; 1, 13, [1, 12]] 1√ 05

π 18

10

√ 12

8

8

+

1

8

=

π 2 11

π 24

√ 5

5

+

deg rad ree ian s s

8

–

2

2√5 5 tan2(18°) = [0; 9, [2,8]]

4 cos2(18°) = [0; 1, 9, [2, 8]] 2 + √2

π 22· 8

√ 8

4

5

=

2

+ 8 cos2(22·5°) = [0; 1, 5, [1, 4]]

8

3

π 30

=

6

√2 – 1 tan(22·5) = [0; [2]] tan2(22·5) = [0; 5, [1,4]]

4

√3

√3

2

3

cos(30°) = [0; 1, 6, [2, 6]] cos2(30°) = [0; 1, 3]

π 36 5

√5

=

60

π 3

tan(30°) = [0; 1, [1,2]] tan2(30°) = [0; 3]

1 +

67· 3π 5 8

3

√5

√ 5 – 2√5 tan2(36°) = [0; 1, 1, [8, 2]]

54 3π 10

4

4

8

8

+

cos(36°) = [0; 1, [4]] 2 cos (36°) = [0; 1, 1, 1, [8, 2]] 5π 37. 24 5

√6 + √3 – √2 – 2 (√2 + 1)(√3 – √2) 1

2

π 45

=

4

52. 7π 5 24

√2

4

1

45

√6 – √3 – √2 + 2 (√2 – 1)(√3 + √2)

52. 5

π 2

cos(45°) = [0; 1, [2]] cos2(45°) = [0; 2] 7π 52. 24 5 1√ 05 3π 54 10

–

1

8

=

π 24

√ 5

5

–

5

8

+

2

2√5 36

π

√3 tan(60°) = [1; [1,2]] tan2(60°) = 3

30

π

1 + √2 tan(67·5) = [2; [2]] tan2(67·5) = [5; [1,4]]

22· 5

5

5

tan2(54°) = [1; 1, [8, 2]] 4 cos2(54°) = [0; 2, 1, [8, 2]] 1

1

π 60

=

3

2

4 cos(60°) = [0; 2] cos2(60°) = [0; 4] 2 – √2

3π 67· 8 5

=

2

4

– 8 2 cos (67·5) = [0; 6, [1, 4]]

√5 2π 72 5

√ 8

4

1 –

4

8

3 = 8 cos(72°) = [0; 3, [4]] cos2(72°) = [0; 10, [2, 8] ]

√5 –

8

√ 5 + 2√5

18 tan2(72°) = [9; [2, 8]]

6

π 8

π 10

√6 5π 75 12

4

√2 –

1 =

4

√ 3 –

4 =

√ 12 –

2 4 cos2(75°) = [0; 14, [1, 12]]

8

8

11

π 82. 5

24

π 90

0

2

2 + √3 tan(75°) = [3; [1, 2]] tan2(75°) = [13; [1,12]]

15

√6 + √3 + √2 + 2 (√2 + 1)(√3 + √2)

7.5

Infinity

0

π 12

π 24 0

The values in the table are those angles of the form n° or n/2 or n/3 for a whole number n, between 0 and 90° whose sin or cosine is rational, or whose continued fraction is periodic or the square of the trig value has a periodic continued fraction. 1

Continued fraction [a; b,c,d,...] means a + 1 b+ 1 c+ d + ...

and the periodic continued fraction [a;b, c, d, e, d, e, d, e, d, e,...] is written as [a; b, c, [d, e]].

Trig functions of Angles <0 or >90° Top of Form

To find the trig. values of all angles including those bigger than 90 degrees and negative angles: 1. select a trig function 2. type the angle in the box and then 3.

click on the button

to find which angle in the range 0-90° has the same value: Select a trig function:

(

°)

Degrees-Radians Converter To convert between DEGREES and RADIANS:

1. enter the angle as a number in one box leaving the other empty 2.

then click the button to do the conversion

You can use Pi in the radians box and * for multiplication e.g. 3*Pi/2: degrees

radians

Bottom of Form

Patterns The Simple Square-Root pattern Ernesto La Orden of Madrid pointed out the following neat way to connect and the easiest of the sines (cosines): Angl e

sine cosine √4

90

30

= 1

2

30

2 √2

1

=

2

√2

√1

1

=

2

2 √0

0

=0

2

The √(2 ± Phi) pattern cos( = 9°)

0

√3

60

45

Angl e

1

2

2 +

2+Φ

45

60

90

1

1

2

2

cos( = 18°)

=

2

cos( = 27°)

+2 + φ 1

2+Φ 2

2 +

2–φ

1

1

Φ

2

2

2

cos( = 36°)

=

=

2 2+φ

+2 – Φ cos( = 54°)

1 2

2 2–Φ –

1 =

2

2–φ

1 2

cos( = 63°)

2

cos( = 72°)

1 2

2 2+φ –

– 2–φ 1 =

2

2–Φ

φ =

2

1 2

cos( = 81°)

2 – 2+Φ

This pattern uses the identities phi = φ = √2 – Φ and Phi = Φ = √2 + φ together with the half-angle formula for cos(A/2) (see below) starting from cos(36)=Phi/2 and cos(72)=phi/2. The pattern continues with the cosines of 4.5°, 13.5°, etc.

The √(2 ± √u) pattern Ernesto La Orden also put many angles into this pattern:

An gle 90

cosine sine

An gle

√2 – √4

= 0

2 75

0

An gle 72

cosine sine √2 – Phi

15

54

√2 – phi

2

60

= 2

45

= 2

30

36

30

18

√2 + √1

=

1

45

√3

60

2

22. √2 + √2 5

67. 5

2

√2 + √3

75 2

0

√2 + √4

= 1

2

√2 + phi

54 2

√2 + Phi

72 2

√2

2

15

1

22. 5

2

√2 – √0

36 2

2

√2 – √1

18 2

√2 – √3

67. √2 – √2 5

An gle

90

The table on the right has values of u that are Phi2 = 2.618033.. and phi2 = 0.381966..

Proofs 30° 45° and 60° Here are two simple triangles which give us the formulae for the trig values of these three angles:-

This triangle is just a square cut along a diagonal. If the sides are of length 1, the diagonal is length √2. This Here is an equilateral triangle where all sides and all angles are equal (to 60°). If the sides are of length 2, gives the sin, cos and tan of 45°. then when we cut it in half as shown, the two triangles have 60°, 30° and 90° angles with a side of length 1 and a hypotenuse of length 2. The other side is therefore of length √3. So we can read off the sin cos and tan of both 30° and 60°.

36° and 54°, 18° and 72° For 36° and 72° we need some further work based on the geometry of a regular pentagon which has angles of 36° and 72°. If the sides of the pentagon are of length 1, the diagonals are of the golden section number in length Phi where: 1 + √5

Phi = = 1.618033 988.. =

2

1

= 1 +

Phi

The upper triangle with angles 72°, 72° and 36° and sides of lengths 1, Phi and Phi shows the trig values for 18° and 72°. The lower triangle with angles of 36°, 36° and 108° and sides of lengths 1, 1 and Phi shows the trig values of 36° and 54°.

15° and 75° If we take the triangle on the left, we can calculate the length of the third side using the Cosine Formula. If, in a triangle with sides a, b and c, we know both sides b and c and also the angle A between sides b and c then we can compute the length of third side, a, as follows: a2 = b2 + c2 – 2 b c cos(A) For our triangle on the left, the known sides are b=2 and c=2 and the angle between them is A=30°. The length of the third side, the base a, is therefore: a2 = 22 + 22 – 2 x 2 x 2 x cos(30°) = 8 – 4 √3

= 2 (4 – 2 √3) But (√3 – 1) = 3 + 1 – 2 √3 = 4 – 2 √3 and so 2

a2 = 2 (√3 – 1)2 Taking the square-root: a = √2 (√3 – 1) which we can also write as = 2 (√3 – 1) / √2 Using this expressions for a, we can expand the triangle by a factor of √2, to get rid of the denominator. Finally, we put in a line from the top of the triangle to the centre of the base a to make two right-angled triangles. This will halve the side a and cut the triangle into two and gets rid of the factor 2 also. We then arrive at the triangle on the right which shows the sines and cosines of 75° and 15°: Ailles Rectangle

An alternative (easier) method for sine and cosine of 15° and 75° is found in Ailles Rectangle (named after an Ontario high school teacher). It is easy to because it is two (green) 45° right-angled triangles stuck onto the sides of a (white) 30-60-90 triangle and the rectangle completed with a (yellow) 15-75-90 triangle on the hypotenuse of the 30-60-90 triangle as shown here. The 30-60-90 sides are "as usual", namely 1, 2 and √3. From the two 45-45-90 triangles, it is quite easy to see that x is √3/√2 and y is 1/√2 from which we can read off the sines and cosines of 15° and 75°.

Trig Formulae Many symmetries and patterns are apparent in the table. They reflect some underlying identities such as: sin(x) = cos(90° – x) ta n( x) = tan sin(x) = a / h 2(x cos(x) = b / ) + 1 h tan(x) = a / b = cot(x) = b / a

sin2(x) + cos2(x) = 1 sin(x)

1

co t(x )=

cos(x) 1

cos(x) =

tan(x)

sin(x) 1

cot2(x )+1 =

sin2(x)

cos2(x)

If we know the value of a trig function on two angles A and B, we can determine the trig function values of their sum and difference using the following identities: sin( A + B ) = sin(A)cos(B) + cos(A)sin(B) sin( A – B ) = sin(A)cos(B) – cos(A)sin(B) cos( A + B ) = cos(A)cos(B) – sin(A)sin(B) cos( A – B ) = cos(A)cos(B) + sin(A)sin(B)

tan(A) + tan(B)

tan(A + B) =

1 – tan(A) tan(B) tan(A) – tan(B)

tan(A – B) =

1 + tan(A) tan(B)

If the two angles are the same (i.e. A=B) we get the sines and cosines of double the angle. Rearranging those formulae gives the formula for the sin or cosine of half an angle: sin( 2A ) = 2 sin(A) cos(A) cos( 2A ) = 1 – 2 sin2(A) cos( 2A ) = cos2(A) – sin2(A) cos( 2A ) = 2 cos2(A) – 1

A si n

2

1 – cos(A) = 2

A

c o s ta n ( 2 A )

2 tan(A)

1 + cos(A) =

2

2

A

sin(A)

1 – cos(A)

2

1 + cos(A)

sin(A)

1 – tan2(A)

=

t a n

=

=

A diagram to relate many angles and Phi Robert Gray's page on Coordinates for many regular solids has an amazing diagram at the bottom which relates Phi to the angles of 18°, 30°, 36°, 45°, 54°, 60° and 72° according to their 3D coordinates in the solids. Each of those angles is measured from the top most point of the circle when a vertical line is turned through that angle. Each line from the base point meets the circle at a point whose a height is 1 (72°), 1+Phi (60°), 2+Phi (54°), 2+2 Phi (45°), 2+3 Phi (36°), 3+3 Phi (30°) or 3+4 Phi (18°). Do look at his pages for more fascinating information on 120 3D solids, of which we will also explore the most symmetrical 5 on our next page.

Things to do 1. Suppose the origin of the circle is the lowest point and its radius is 2 + 2 Phi. Find the equation of the circle. 2. Use your answer to the previous question to find the coordinates of each of the points on the circle with the angles shown. 3. Compute the lengths of each of the red lines from the lowest point to the points shown on the circle. 4. From any two points A and B on a circle, the angle AOB at the centre of a circle, O, is twice the angle at any point on the circumference in the same sector. In the diagram,all the red angles at the circumference are equal; the red angles are twice the blue angle AOB at the centre; the red angles are to a point in the same sector of the circle as is the centre of the circle so they cannot be in the grey sector.

Use the above theorem to find three points on the circle ABOVE this Things To Do section where a line from the centre makes an angle with the vertical of i. 2×18=36° ii. 2×30=60° iii. 2×36=72°

Other angles with exact trig expressions involving square-roots Are there other angles with a simple exact expression for their cosine or sine? Well it all depends upon what you mean by simple! Carl Friedrich GAUSS (177 - 1855) looked at a similar problem which answers this question. He investigated if there was a method of constructing a regular polygon of n sides using only a pair of comes (to draw circles) and a straight-edge (a ruler with no markings). We know we can construct a regular polygon for all of the values of n=3, 4, 5, 6, 8 and 10. Halving There is a simple geometrical way to use comes to divide an angle into two (angle bisection). So all the angles in a regular n-gon can be split into two to make a regular 2n-gon. We can repeat the process to get a 4n-gon, 8n-gon and in general a 2kn-gon for any k once we have a method of constructing a regular n-gon. The Trig Formula section above contains a formula for the cosine of half an angle in of the cosine of the (whole) angle: c o s

A

1 + cos(A) =

2

2 + 2cos(A) =

2

2

As Mitch Wyatt pointed out to me, since we know that cos(90°) is 0 and 90° is /2 radians, we can use it to find the cosine of half that angle (45° or /4 radians) and then halve that angle again and so on. Each time we introduce another square root so we get a cascading or nested sequence of square roots: co s co

√ 2 4

= 2

s

√2√ 2 = +

8

2 co s

√ 2√2√ 2 + + =

16

2 co s

√

√ 2 2√ 2√ 2 = +++

32

2

However, this page is about sines and cosines which have simpler expressions, so we will not expand on this except to say that it shows how we can always find an exact expression for the sine (or cosine) of 1/2, 1/4, 1/8, ..., 1/2 , ... of any angle for which we have an exact sine (or cosine) expression. n

Superimposing If we construct a regular triangle (3 sides) and with the same circle centre, construct three regular pentagons (5 sides) with each having one vertex in common with the triangle, we will have the 15 vertices of a regular 15-gon. This is shown on the right with the 3 pentagons in blue on the same circle, each having a vertex in common with the red triangle and the regular 15-gon appears in yellow. By superimposing two regular polygons like this, we can construct a regular P×Q-gon (if P and Q have no factors in common otherwise more than one vertex of each will coincide). Do we know all the angles? All this was known in Euclid's time, around the year 300 BC. So what about 7ths and 9ths? Is it possible to find sines and cosines of all the multiples of 1/7 and 1/9 of a turn in exact (using square roots)? What about 11ths and 12ths etc.? In the next 2000 years no one found an exact geometric method for 7-gons or 9-gons but also no one had proved it was impossible to construct such regular polygons. Then C F Gauss completely solved the problem while he was a student at Göttingen between 1795 and 1798. Gauss found the conditions on n and its the prime factors to solve two equivalent problems: • drawing a regular n-sided polygon using only a straight edge and com and •

expressing the cos and sin of 360/n° using only square roots. If we factor n as 2ap1bp2c..., i.e. a, b, c, ... are the powers of n's prime factors: 2, p1, p2, ... (the prime's power is 0 if it is not a factor of n) then both of the problems are solvable when

•

b,c,... and all the powers except a, the power of 2, must be 1, and

•

the primes>2 that are factors of n (that is p1, p2, ...) must be of the form 22 +1 for some number k. k

Both problems are solvable for these values of n and only for these values. Prime numbers of the form 22 +1 are called Fermat primes. The series of numbers of the form 22 +1 begins k

k

22 + 1 = 3, 22 + 1 = 5, 22 + 1 = 17, 22 + 1 = 257, 22 + 1 = 65537, ... However not every number of the form 22 + 1 is prime -- and it is only the prime ones that we must have as factors of n. The next one, 22 + 1 is 4294967297 and has a factor of 641 so it is not prime. In fact, we do not know if there are any more primes of this form except the first 5 listed above. 0

1

2

3

4

k

5

Such numbers, n, of the form Gauss describes are as follows, one per row, each a product of some of the Fermat primes (but each prime at most once) followed by its multiples of two. For any number in the table, its double is also in the table: 24 36 5 10 3×5=15 30 17 34 3×17=51 102 5×17=75 150 3×5×17=225 450 257 514

8 12 20 60 68 204 300 900 1028

16 24 40 120 136 408 600 1800 2056

32 48 80 240 ... ... ... ... ...

... ... ... ...

When put in order, we have (1), 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, ... which is Sloane's A003401. Here, for instance, is the cosine of an angle involving 17ths:

8

co s

√1 √ 7 – 68 + 12√17 + 2(√17 – 1)S – 16 T 1 + =S –

17 wher√ e S 34 – 2√17 =

16 an √ d T 34 + 2√17 =

Things to do 1.

From the five 'starting values' above: 3, 5, 17, 257 and 65537 we can multiply these and also double any number to get a constructible polygon or an angle with a sine which involves nothing more than square-roots. Don't forget that we can double 2 to get 4 (a square), 8 (octagon), 16, etc too but we cannot use any of the 5 odd primes more than once in any product (so 3x3=9 and 5x5=25 are not in the list but 3x5x7 is). A complete list will involve these five numbers of course together with their products 3x5=15, 3x17=51, ... and you can double any of these any number of times: 2x3=6, 2x15=30 as well as 2x6=12, 2x12=24,... . a.

What are the first 12 values in the list that starts 3, 4, 5, ...?

b. Check that there are 24 values (excluding 1 and 2) less than 100. c. 2.

Is 100 in the list?

From the five known values: 3, 5, 17, 257 and 65537, there are a finite number of odd numbers n for which sin(360/n°) can be written with square-roots alone i.e. products of these 5 numbers where no number can be used more than once. How many odd numbers (excluding 1) can you make using these 5 no more than once in each product? This is the total number of known polygons we can construct with ruler and com or which have a sine (cosine) formed from nothing more than square-roots. With thanks to Richard Duffy for suggesting this puzzle

Tom Ace pointed out that there is more about this in chapter 15 of Oystein Ore's Number Theory and Its History from 1948 and now available as a Dover book(1988). What angles have an exact expression for their sines, cosines and tangents? You might know that cos(60°)=1/2 and sin(60°)=√3/2 as well as tan(45°)=1, but are 30, 45 and 60 the only angles up to 90° with a formula for their trig values? No! There are lots more but not all angles have exact epxressions. Which angles do? What patterns are there in these expressions? This page shows expressions for many angles and even solves the complete problem of which angles do and which don't have exact trig expressions.

Contents of this Page The

icon means there is a Things to do investigation at the end of the section.

The sections marked with •

have an online interactive calculator.

A Table of Exact Trig values 

Converter for trig functions of angles <0 or >90°

•

Degress <-> Radians Converter

•

Patterns

•



The Simple Square-Root pattern



The √(2 ± Phi) pattern



The √(2 ± √u) pattern

Proofs 

30° 45° and 60°



36° and 54°, 18° and 72°



15° and 75° ○

Ailles Rectangle

•

Trig Formulae

•

A diagram to relate many angles and Phi

•

Other angles with exact trig expressions involving square-roots



Halving



Superimposing



Do we know all the angles?

A Table of Exact Trig values that are expressible as simple involving square-roots. a

b

rad deg ian ree s s

0

0

cos(a) sin(b)

tan(a) cot(b)

1

0

90

√6 – √3 + √2 – 2 (√2 – 1)(√3 – √2)

82. 5

2 – √3 tan(15°) = [0; 3, [1, 2]] tan2(15°) = [0; 13, [1, 12]]

75

5π 12

72

2π 5

π 7.5

24

√6

π 15

4

12

√2 +

1 4

=

√ 3 +

4 =

+

2 4 cos2(15°) = [0; 1, 13, [1, 12]] 1√ 05

π 18

10

5

√ 12

8

8

=

π 2 11

π 24

√ 5 +

+

deg rad ree ian s s

1

8

8

2

–

2√5 5 tan2(18°) = [0; 9, [2,8]]

4 cos2(18°) = [0; 1, 9, [2, 8]] 2 + √2

4

π 22· 8

5

π 30 6

=

2

√ 8 +

8 cos2(22·5°) = [0; 1, 5, [1, 4]] √3

3 = 4

8

√2 – 1 tan(22·5) = [0; [2]] tan2(22·5) = [0; 5, [1,4]]

√3

67· 3π 5 8

60

π 3

2

3

cos(30°) = [0; 1, 6, [2, 6]] cos2(30°) = [0; 1, 3] √5

π 36

1 +

4

5

4

tan(30°) = [0; 1, [1,2]] tan2(30°) = [0; 3] 3

√5

=

+

8

8

√ 5 – 2√5 tan (36°) = [0; 1, 1, [8, 2]]

cos(36°) = [0; 1, [4]] cos2(36°) = [0; 1, 1, 1, [8, 2]] 5π 37. 24 5

√6 + √3 – √2 – 2 (√2 + 1)(√3 – √2)

=

4

3π 10

52. 7π 5 24

1

2

π 45

54

2

√2

4

1

45

√6 – √3 – √2 + 2 (√2 – 1)(√3 + √2)

52. 5

π 2

cos(45°) = [0; 1, [2]] cos2(45°) = [0; 2] 7π 52. 24 5 1√ 05 3π 54 10

–

1

8

=

π 24

√ 5

5

–

5

8

+

2

2√5 36

π

√3 tan(60°) = [1; [1,2]] tan2(60°) = 3

30

π

1 + √2 tan(67·5) = [2; [2]] tan2(67·5) = [5; [1,4]]

22· 5

π

18

π

5

5

tan2(54°) = [1; 1, [8, 2]] 4 cos2(54°) = [0; 2, 1, [8, 2]] 1

1

π 60

=

3

2

4 cos(60°) = [0; 2] cos2(60°) = [0; 4] 2 – √2

3π 67· 8 5

2π 72 5

√ 8

4 =

2

– 8 cos2(67·5) = [0; 6, [1, 4]]

√5

1 –

=

8

3

√5

√ 5 + 2√5

6

8

10

4

4

8

8

–

tan2(72°) = [9; [2, 8]]

cos(72°) = [0; 3, [4]] cos2(72°) = [0; 10, [2, 8] ] √6 5π 75 12

4

√2 –

1 =

4

√ 3 –

4 =

√ 12 –

2 4 2 cos (75°) = [0; 14, [1, 12]]

8

8

11

π 82. 5

24

π 90

0

2

2 + √3 tan(75°) = [3; [1, 2]] tan2(75°) = [13; [1,12]]

15

√6 + √3 + √2 + 2 (√2 + 1)(√3 + √2)

7.5

Infinity

0

π 12

π 24 0

The values in the table are those angles of the form n° or n/2 or n/3 for a whole number n, between 0 and 90° whose sin or cosine is rational, or whose continued fraction is periodic or the square of the trig value has a periodic continued fraction. 1

Continued fraction [a; b,c,d,...] means a + 1 b+ 1 c+ d + ...

and the periodic continued fraction [a;b, c, d, e, d, e, d, e, d, e,...] is written as [a; b, c, [d, e]].

Trig functions of Angles <0 or >90° Top of Form

To find the trig. values of all angles including those bigger than 90 degrees and negative angles: 1. select a trig function 2. type the angle in the box and then 3.

click on the button

to find which angle in the range 0-90° has the same value: Select a trig function:

(

°)

Degrees-Radians Converter To convert between DEGREES and RADIANS: 1. enter the angle as a number in one box leaving the other empty 2.

then click the button to do the conversion

You can use Pi in the radians box and * for multiplication e.g. 3*Pi/2: degrees

radians

Bottom of Form

Patterns The Simple Square-Root pattern Ernesto La Orden of Madrid pointed out the following neat way to connect and the easiest of the sines (cosines): Angl e

sine cosine √4

90

30

= 1

2

30

2 √2

1

=

2

√2

√1

1

=

2

2 √0

0

=0

2

The √(2 ± Phi) pattern cos( = 9°)

0

√3

60

45

Angl e

1

2

2 +

2+Φ

45

60

90

1

1

2

2

cos( = 18°)

=

2

cos( = 27°)

+2 + φ 1

2+Φ 2

2 +

2–φ

1

1

Φ

2

2

2

cos( = 36°)

=

=

2 2+φ

+2 – Φ cos( = 54°)

1 2

2 2–Φ –

1 =

2

2–φ

1 2

cos( = 63°)

2

cos( = 72°)

1 2

2 2+φ –

– 2–φ 1 =

2

2–Φ

φ =

2

1 2

cos( = 81°)

2 – 2+Φ

This pattern uses the identities phi = φ = √2 – Φ and Phi = Φ = √2 + φ together with the half-angle formula for cos(A/2) (see below) starting from cos(36)=Phi/2 and cos(72)=phi/2. The pattern continues with the cosines of 4.5°, 13.5°, etc.

The √(2 ± √u) pattern Ernesto La Orden also put many angles into this pattern:

An gle 90

cosine sine

An gle

√2 – √4

= 0

2 75

0

An gle 72

cosine sine √2 – Phi

15

54

√2 – phi

2

60

= 2

45

= 2

30

36

30

18

√2 + √1

=

1

45

√3

60

2

22. √2 + √2 5

67. 5

2

√2 + √3

75 2

0

√2 + √4

= 1

2

√2 + phi

54 2

√2 + Phi

72 2

√2

2

15

1

22. 5

2

√2 – √0

36 2

2

√2 – √1

18 2

√2 – √3

67. √2 – √2 5

An gle

90

The table on the right has values of u that are Phi2 = 2.618033.. and phi2 = 0.381966..

Proofs 30° 45° and 60° Here are two simple triangles which give us the formulae for the trig values of these three angles:-

This triangle is just a square cut along a diagonal. If the sides are of length 1, the diagonal is length √2. This Here is an equilateral triangle where all sides and all angles are equal (to 60°). If the sides are of length 2, gives the sin, cos and tan of 45°. then when we cut it in half as shown, the two triangles have 60°, 30° and 90° angles with a side of length 1 and a hypotenuse of length 2. The other side is therefore of length √3. So we can read off the sin cos and tan of both 30° and 60°.

36° and 54°, 18° and 72° For 36° and 72° we need some further work based on the geometry of a regular pentagon which has angles of 36° and 72°. If the sides of the pentagon are of length 1, the diagonals are of the golden section number in length Phi where: 1 + √5

Phi = = 1.618033 988.. =

2

1

= 1 +

Phi

The upper triangle with angles 72°, 72° and 36° and sides of lengths 1, Phi and Phi shows the trig values for 18° and 72°. The lower triangle with angles of 36°, 36° and 108° and sides of lengths 1, 1 and Phi shows the trig values of 36° and 54°.

15° and 75° If we take the triangle on the left, we can calculate the length of the third side using the Cosine Formula. If, in a triangle with sides a, b and c, we know both sides b and c and also the angle A between sides b and c then we can compute the length of third side, a, as follows: a2 = b2 + c2 – 2 b c cos(A) For our triangle on the left, the known sides are b=2 and c=2 and the angle between them is A=30°. The length of the third side, the base a, is therefore: a2 = 22 + 22 – 2 x 2 x 2 x cos(30°) = 8 – 4 √3

= 2 (4 – 2 √3) But (√3 – 1) = 3 + 1 – 2 √3 = 4 – 2 √3 and so 2

a2 = 2 (√3 – 1)2 Taking the square-root: a = √2 (√3 – 1) which we can also write as = 2 (√3 – 1) / √2 Using this expressions for a, we can expand the triangle by a factor of √2, to get rid of the denominator. Finally, we put in a line from the top of the triangle to the centre of the base a to make two right-angled triangles. This will halve the side a and cut the triangle into two and gets rid of the factor 2 also. We then arrive at the triangle on the right which shows the sines and cosines of 75° and 15°: Ailles Rectangle

An alternative (easier) method for sine and cosine of 15° and 75° is found in Ailles Rectangle (named after an Ontario high school teacher). It is easy to because it is two (green) 45° right-angled triangles stuck onto the sides of a (white) 30-60-90 triangle and the rectangle completed with a (yellow) 15-75-90 triangle on the hypotenuse of the 30-60-90 triangle as shown here. The 30-60-90 sides are "as usual", namely 1, 2 and √3. From the two 45-45-90 triangles, it is quite easy to see that x is √3/√2 and y is 1/√2 from which we can read off the sines and cosines of 15° and 75°.

Trig Formulae Many symmetries and patterns are apparent in the table. They reflect some underlying identities such as: sin(x) = cos(90° – x) ta n( x) = tan sin(x) = a / h 2(x cos(x) = b / ) + 1 h tan(x) = a / b = cot(x) = b / a

sin2(x) + cos2(x) = 1 sin(x)

1

co t(x )=

cos(x) 1

cos(x) =

tan(x)

sin(x) 1

cot2(x )+1 =

sin2(x)

cos2(x)

If we know the value of a trig function on two angles A and B, we can determine the trig function values of their sum and difference using the following identities: sin( A + B ) = sin(A)cos(B) + cos(A)sin(B) sin( A – B ) = sin(A)cos(B) – cos(A)sin(B) cos( A + B ) = cos(A)cos(B) – sin(A)sin(B) cos( A – B ) = cos(A)cos(B) + sin(A)sin(B)

tan(A) + tan(B)

tan(A + B) =

1 – tan(A) tan(B) tan(A) – tan(B)

tan(A – B) =

1 + tan(A) tan(B)

If the two angles are the same (i.e. A=B) we get the sines and cosines of double the angle. Rearranging those formulae gives the formula for the sin or cosine of half an angle: sin( 2A ) = 2 sin(A) cos(A) cos( 2A ) = 1 – 2 sin2(A) cos( 2A ) = cos2(A) – sin2(A) cos( 2A ) = 2 cos2(A) – 1

A si n

2

1 – cos(A) = 2

A

c o s ta n ( 2 A )

2 tan(A)

1 + cos(A) =

2

2

A

sin(A)

1 – cos(A)

2

1 + cos(A)

sin(A)

1 – tan2(A)

=

t a n

=

=

A diagram to relate many angles and Phi Robert Gray's page on Coordinates for many regular solids has an amazing diagram at the bottom which relates Phi to the angles of 18°, 30°, 36°, 45°, 54°, 60° and 72° according to their 3D coordinates in the solids. Each of those angles is measured from the top most point of the circle when a vertical line is turned through that angle. Each line from the base point meets the circle at a point whose a height is 1 (72°), 1+Phi (60°), 2+Phi (54°), 2+2 Phi (45°), 2+3 Phi (36°), 3+3 Phi (30°) or 3+4 Phi (18°). Do look at his pages for more fascinating information on 120 3D solids, of which we will also explore the most symmetrical 5 on our next page.

Things to do 1. Suppose the origin of the circle is the lowest point and its radius is 2 + 2 Phi. Find the equation of the circle. 2. Use your answer to the previous question to find the coordinates of each of the points on the circle with the angles shown. 3. Compute the lengths of each of the red lines from the lowest point to the points shown on the circle. 4. From any two points A and B on a circle, the angle AOB at the centre of a circle, O, is twice the angle at any point on the circumference in the same sector. In the diagram,all the red angles at the circumference are equal; the red angles are twice the blue angle AOB at the centre; the red angles are to a point in the same sector of the circle as is the centre of the circle so they cannot be in the grey sector.

Use the above theorem to find three points on the circle ABOVE this Things To Do section where a line from the centre makes an angle with the vertical of i. 2×18=36° ii. 2×30=60° iii. 2×36=72°

Other angles with exact trig expressions involving square-roots Are there other angles with a simple exact expression for their cosine or sine? Well it all depends upon what you mean by simple! Carl Friedrich GAUSS (177 - 1855) looked at a similar problem which answers this question. He investigated if there was a method of constructing a regular polygon of n sides using only a pair of comes (to draw circles) and a straight-edge (a ruler with no markings). We know we can construct a regular polygon for all of the values of n=3, 4, 5, 6, 8 and 10. Halving There is a simple geometrical way to use comes to divide an angle into two (angle bisection). So all the angles in a regular n-gon can be split into two to make a regular 2n-gon. We can repeat the process to get a 4n-gon, 8n-gon and in general a 2kn-gon for any k once we have a method of constructing a regular n-gon. The Trig Formula section above contains a formula for the cosine of half an angle in of the cosine of the (whole) angle: c o s

A

1 + cos(A) =

2

2 + 2cos(A) =

2

2

As Mitch Wyatt pointed out to me, since we know that cos(90°) is 0 and 90° is /2 radians, we can use it to find the cosine of half that angle (45° or /4 radians) and then halve that angle again and so on. Each time we introduce another square root so we get a cascading or nested sequence of square roots: co s co

√ 2 4

= 2

s

√2√ 2 = +

8

2 co s

√ 2√2√ 2 + + =

16

2 co s

√

√ 2 2√ 2√ 2 = +++

32

2

However, this page is about sines and cosines which have simpler expressions, so we will not expand on this except to say that it shows how we can always find an exact expression for the sine (or cosine) of 1/2, 1/4, 1/8, ..., 1/2 , ... of any angle for which we have an exact sine (or cosine) expression. n

Superimposing If we construct a regular triangle (3 sides) and with the same circle centre, construct three regular pentagons (5 sides) with each having one vertex in common with the triangle, we will have the 15 vertices of a regular 15-gon. This is shown on the right with the 3 pentagons in blue on the same circle, each having a vertex in common with the red triangle and the regular 15-gon appears in yellow. By superimposing two regular polygons like this, we can construct a regular P×Q-gon (if P and Q have no factors in common otherwise more than one vertex of each will coincide). Do we know all the angles? All this was known in Euclid's time, around the year 300 BC. So what about 7ths and 9ths? Is it possible to find sines and cosines of all the multiples of 1/7 and 1/9 of a turn in exact (using square roots)? What about 11ths and 12ths etc.? In the next 2000 years no one found an exact geometric method for 7-gons or 9-gons but also no one had proved it was impossible to construct such regular polygons. Then C F Gauss completely solved the problem while he was a student at Göttingen between 1795 and 1798. Gauss found the conditions on n and its the prime factors to solve two equivalent problems: • drawing a regular n-sided polygon using only a straight edge and com and •

expressing the cos and sin of 360/n° using only square roots. If we factor n as 2ap1bp2c..., i.e. a, b, c, ... are the powers of n's prime factors: 2, p1, p2, ... (the prime's power is 0 if it is not a factor of n) then both of the problems are solvable when

•

b,c,... and all the powers except a, the power of 2, must be 1, and

•

the primes>2 that are factors of n (that is p1, p2, ...) must be of the form 22 +1 for some number k. k

Both problems are solvable for these values of n and only for these values. Prime numbers of the form 22 +1 are called Fermat primes. The series of numbers of the form 22 +1 begins k

k

22 + 1 = 3, 22 + 1 = 5, 22 + 1 = 17, 22 + 1 = 257, 22 + 1 = 65537, ... However not every number of the form 22 + 1 is prime -- and it is only the prime ones that we must have as factors of n. The next one, 22 + 1 is 4294967297 and has a factor of 641 so it is not prime. In fact, we do not know if there are any more primes of this form except the first 5 listed above. 0

1

2

3

4

k

5

Such numbers, n, of the form Gauss describes are as follows, one per row, each a product of some of the Fermat primes (but each prime at most once) followed by its multiples of two. For any number in the table, its double is also in the table: 24 36 5 10 3×5=15 30 17 34 3×17=51 102 5×17=75 150 3×5×17=225 450 257 514

8 12 20 60 68 204 300 900 1028

16 24 40 120 136 408 600 1800 2056

32 48 80 240 ... ... ... ... ...

... ... ... ...

When put in order, we have (1), 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, ... which is Sloane's A003401. Here, for instance, is the cosine of an angle involving 17ths:

8

co s

√1 √ 7 – 68 + 12√17 + 2(√17 – 1)S – 16 T 1 + =S –

17 wher√ e S 34 – 2√17 =

16 an √ d T 34 + 2√17 =

Things to do 1.

From the five 'starting values' above: 3, 5, 17, 257 and 65537 we can multiply these and also double any number to get a constructible polygon or an angle with a sine which involves nothing more than square-roots. Don't forget that we can double 2 to get 4 (a square), 8 (octagon), 16, etc too but we cannot use any of the 5 odd primes more than once in any product (so 3x3=9 and 5x5=25 are not in the list but 3x5x7 is). A complete list will involve these five numbers of course together with their products 3x5=15, 3x17=51, ... and you can double any of these any number of times: 2x3=6, 2x15=30 as well as 2x6=12, 2x12=24,... . a.

What are the first 12 values in the list that starts 3, 4, 5, ...?

b. Check that there are 24 values (excluding 1 and 2) less than 100. c. 2.

Is 100 in the list?

From the five known values: 3, 5, 17, 257 and 65537, there are a finite number of odd numbers n for which sin(360/n°) can be written with square-roots alone i.e. products of these 5 numbers where no number can be used more than once. How many odd numbers (excluding 1) can you make using these 5 no more than once in each product? This is the total number of known polygons we can construct with ruler and com or which have a sine (cosine) formed from nothing more than square-roots. With thanks to Richard Duffy for suggesting this puzzle

Tom Ace pointed out that there is more about this in chapter 15 of Oystein Ore's Number Theory and Its History from 1948 and now available as a Dover book(1988). If you are printing this page and the horizontal lines of fractions or the square-roots do not appear, make sure you have the "Print Images" and "Print Background" options checked when your browser's printing window appears. What angles have an exact expression for their sines, cosines and tangents? You might know that cos(60°)=1/2 and sin(60°)=√3/2 as well as tan(45°)=1, but are 30, 45 and 60 the only angles up to 90° with a formula for their trig values? No! There are lots more but not all angles have exact epxressions. Which angles do? What patterns are there in these expressions? This page shows expressions for many angles and even solves the complete problem of which angles do and which don't have exact trig expressions.

Contents of this Page The

icon means there is a Things to do investigation at the end of the section.

The sections marked with •

have an online interactive calculator.

A Table of Exact Trig values 

Converter for trig functions of angles <0 or >90°

•

Degress <-> Radians Converter

•

Patterns

•



The Simple Square-Root pattern



The √(2 ± Phi) pattern



The √(2 ± √u) pattern

Proofs 

30° 45° and 60°



36° and 54°, 18° and 72°



15° and 75° ○

•

Ailles Rectangle

Trig Formulae

•

A diagram to relate many angles and Phi

•

Other angles with exact trig expressions involving square-roots 

Halving



Superimposing



Do we know all the angles?

A Table of Exact Trig values that are expressible as simple involving square-roots. a

b

rad deg ian ree s s

0

0

cos(a) sin(b)

tan(a) cot(b)

1

0

90

√6 – √3 + √2 – 2 (√2 – 1)(√3 – √2)

82. 5

2 – √3 tan(15°) = [0; 3, [1, 2]] tan2(15°) = [0; 13, [1, 12]]

75

5π 12

72

2π 5

π 7.5

24

√6

π 15

4

12

√2 +

√ 3

1 4

=

+

4 =

+

2 4 cos2(15°) = [0; 1, 13, [1, 12]] 1√ 05

5

8

8

1

8

=

π 2 11

π 24

√ 5 +

π 18 +

10

√ 12

deg rad ree ian s s

8

2

–

2√5 5 tan2(18°) = [0; 9, [2,8]]

4 cos2(18°) = [0; 1, 9, [2, 8]] 2 + √2

4

π 22· 8

5

=

2

√ 8 +

8 cos (22·5°) = [0; 1, 5, [1, 4]]

8

√2 – 1 tan(22·5) = [0; [2]] tan2(22·5) = [0; 5, [1,4]]

67· 3π 5 8

2

π 30

3

√3

√3

60

π

=

4

6

3

2

cos(30°) = [0; 1, 6, [2, 6]] cos2(30°) = [0; 1, 3] √5

π 36

1 +

4

5

4

tan(30°) = [0; 1, [1,2]] tan2(30°) = [0; 3] 3

√5

=

+

8

8

√ 5 – 2√5 tan2(36°) = [0; 1, 1, [8, 2]]

cos(36°) = [0; 1, [4]] 2 cos (36°) = [0; 1, 1, 1, [8, 2]] 5π 37. 24 5

√6 + √3 – √2 – 2 (√2 + 1)(√3 – √2)

=

4

54

3π 10

52. 7π 5 24

1

2

π 45

3

√2

4

1

45

√6 – √3 – √2 + 2 (√2 – 1)(√3 + √2)

52. 5

π 2

cos(45°) = [0; 1, [2]] cos2(45°) = [0; 2] 7π 52. 24 5 1√ 05 3π 54 10

– =

π 24

√ 5

5

–

5

1

8

8

2

+

2√5 5

36

π

30

π

5

tan2(54°) = [1; 1, [8, 2]] 4 cos2(54°) = [0; 2, 1, [8, 2]] 1

1

π 60

=

3

√3 tan(60°) = [1; [1,2]] tan2(60°) = 3

2

4 cos(60°) = [0; 2] cos2(60°) = [0; 4]

3π 67· 2 – √2 8 5

4 2

=

1 + √2 tan(67·5) = [2; [2]] tan2(67·5) = [5; [1,4]]

√ 8 –

8

8

22· 5

6

π 8

cos2(67·5) = [0; 6, [1, 4]] √5 2π 72 5

1 –

4

3

√5

=

4

–

8

√ 5 + 2√5

8

tan2(72°) = [9; [2, 8]]

cos(72°) = [0; 3, [4]] cos2(72°) = [0; 10, [2, 8] ] √6 5π 75 12

4

√2 –

1 =

4

√ 3 –

4 =

2 4 cos2(75°) = [0; 14, [1, 12]]

√ 12 –

8

8

11

π 82. 5

24

π 90

0

2

18

2 + √3 tan(75°) = [3; [1, 2]] tan2(75°) = [13; [1,12]]

15

√6 + √3 + √2 + 2 (√2 + 1)(√3 + √2)

7.5

Infinity

0

π 10

π 12

π 24 0

The values in the table are those angles of the form n° or n/2 or n/3 for a whole number n, between 0 and 90° whose sin or cosine is rational, or whose continued fraction is periodic or the square of the trig value has a periodic continued fraction. 1

Continued fraction [a; b,c,d,...] means a + 1 b+ 1 c+ d + ...

and the periodic continued fraction [a;b, c, d, e, d, e, d, e, d, e,...] is written as [a; b, c, [d, e]].

Trig functions of Angles <0 or >90° Top of Form

To find the trig. values of all angles including those bigger than 90 degrees and negative angles: 1. select a trig function 2. type the angle in the box and then 3.

click on the button

to find which angle in the range 0-90° has the same value:

Select a trig function:

(

°)

Degrees-Radians Converter To convert between DEGREES and RADIANS: 1. enter the angle as a number in one box leaving the other empty 2.

then click the button to do the conversion

You can use Pi in the radians box and * for multiplication e.g. 3*Pi/2: degrees

radians

Bottom of Form

Patterns The Simple Square-Root pattern Ernesto La Orden of Madrid pointed out the following neat way to connect and the easiest of the sines (cosines): Angl e

sine cosine √4

90

30

= 1

2

30

2 √2

1

=

2

√2

√1

1

=

2

2 √0

0

=0

2

The √(2 ± Phi) pattern cos( = 9°)

0

√3

60

45

Angl e

1

2

2 +

2+Φ

45

60

90

1

1

2

2

cos( = 18°)

=

2

cos( = 27°)

+2 + φ 1

2+Φ 2

2 +

2–φ

1

1

Φ

2

2

2

cos( = 36°)

=

=

2 2+φ

+2 – Φ cos( = 54°)

1 2

2 2–Φ –

1 =

2

2–φ

1 2

cos( = 63°)

2

cos( = 72°)

1 2

2 2+φ –

– 2–φ 1 =

2

2–Φ

φ =

2

1 2

cos( = 81°)

2 – 2+Φ

This pattern uses the identities phi = φ = √2 – Φ and Phi = Φ = √2 + φ together with the half-angle formula for cos(A/2) (see below) starting from cos(36)=Phi/2 and cos(72)=phi/2. The pattern continues with the cosines of 4.5°, 13.5°, etc.

The √(2 ± √u) pattern Ernesto La Orden also put many angles into this pattern:

An gle 90

cosine sine

An gle

√2 – √4

= 0

2 75

0

An gle 72

cosine sine √2 – Phi

15

54

√2 – phi

2

60

= 2

45

= 2

30

36

30

18

√2 + √1

=

1

45

√3

60

2

22. √2 + √2 5

67. 5

2

√2 + √3

75 2

0

√2 + √4

= 1

2

√2 + phi

54 2

√2 + Phi

72 2

√2

2

15

1

22. 5

2

√2 – √0

36 2

2

√2 – √1

18 2

√2 – √3

67. √2 – √2 5

An gle

90

The table on the right has values of u that are Phi2 = 2.618033.. and phi2 = 0.381966..

Proofs 30° 45° and 60° Here are two simple triangles which give us the formulae for the trig values of these three angles:-

This triangle is just a square cut along a diagonal. If the sides are of length 1, the diagonal is length √2. This Here is an equilateral triangle where all sides and all angles are equal (to 60°). If the sides are of length 2, gives the sin, cos and tan of 45°. then when we cut it in half as shown, the two triangles have 60°, 30° and 90° angles with a side of length 1 and a hypotenuse of length 2. The other side is therefore of length √3. So we can read off the sin cos and tan of both 30° and 60°.

36° and 54°, 18° and 72° For 36° and 72° we need some further work based on the geometry of a regular pentagon which has angles of 36° and 72°. If the sides of the pentagon are of length 1, the diagonals are of the golden section number in length Phi where: 1 + √5

Phi = = 1.618033 988.. =

2

1

= 1 +

Phi

The upper triangle with angles 72°, 72° and 36° and sides of lengths 1, Phi and Phi shows the trig values for 18° and 72°. The lower triangle with angles of 36°, 36° and 108° and sides of lengths 1, 1 and Phi shows the trig values of 36° and 54°.

15° and 75° If we take the triangle on the left, we can calculate the length of the third side using the Cosine Formula. If, in a triangle with sides a, b and c, we know both sides b and c and also the angle A between sides b and c then we can compute the length of third side, a, as follows: a2 = b2 + c2 – 2 b c cos(A) For our triangle on the left, the known sides are b=2 and c=2 and the angle between them is A=30°. The length of the third side, the base a, is therefore: a2 = 22 + 22 – 2 x 2 x 2 x cos(30°) = 8 – 4 √3

= 2 (4 – 2 √3) But (√3 – 1) = 3 + 1 – 2 √3 = 4 – 2 √3 and so 2

a2 = 2 (√3 – 1)2 Taking the square-root: a = √2 (√3 – 1) which we can also write as = 2 (√3 – 1) / √2 Using this expressions for a, we can expand the triangle by a factor of √2, to get rid of the denominator. Finally, we put in a line from the top of the triangle to the centre of the base a to make two right-angled triangles. This will halve the side a and cut the triangle into two and gets rid of the factor 2 also. We then arrive at the triangle on the right which shows the sines and cosines of 75° and 15°: Ailles Rectangle

An alternative (easier) method for sine and cosine of 15° and 75° is found in Ailles Rectangle (named after an Ontario high school teacher). It is easy to because it is two (green) 45° right-angled triangles stuck onto the sides of a (white) 30-60-90 triangle and the rectangle completed with a (yellow) 15-75-90 triangle on the hypotenuse of the 30-60-90 triangle as shown here. The 30-60-90 sides are "as usual", namely 1, 2 and √3. From the two 45-45-90 triangles, it is quite easy to see that x is √3/√2 and y is 1/√2 from which we can read off the sines and cosines of 15° and 75°.

Trig Formulae Many symmetries and patterns are apparent in the table. They reflect some underlying identities such as: sin(x) = cos(90° – x) ta n( x) = tan sin(x) = a / h 2(x cos(x) = b / ) + 1 h tan(x) = a / b = cot(x) = b / a

sin2(x) + cos2(x) = 1 sin(x)

1

co t(x )=

cos(x) 1

cos(x) =

tan(x)

sin(x) 1

cot2(x )+1 =

sin2(x)

cos2(x)

If we know the value of a trig function on two angles A and B, we can determine the trig function values of their sum and difference using the following identities: sin( A + B ) = sin(A)cos(B) + cos(A)sin(B) sin( A – B ) = sin(A)cos(B) – cos(A)sin(B) cos( A + B ) = cos(A)cos(B) – sin(A)sin(B) cos( A – B ) = cos(A)cos(B) + sin(A)sin(B)

tan(A) + tan(B)

tan(A + B) =

1 – tan(A) tan(B) tan(A) – tan(B)

tan(A – B) =

1 + tan(A) tan(B)

If the two angles are the same (i.e. A=B) we get the sines and cosines of double the angle. Rearranging those formulae gives the formula for the sin or cosine of half an angle: sin( 2A ) = 2 sin(A) cos(A) cos( 2A ) = 1 – 2 sin2(A) cos( 2A ) = cos2(A) – sin2(A) cos( 2A ) = 2 cos2(A) – 1

A si n

2

1 – cos(A) = 2

A

c o s ta n ( 2 A )

2 tan(A)

1 + cos(A) =

2

2

A

sin(A)

1 – cos(A)

2

1 + cos(A)

sin(A)

1 – tan2(A)

=

t a n

=

=

A diagram to relate many angles and Phi Robert Gray's page on Coordinates for many regular solids has an amazing diagram at the bottom which relates Phi to the angles of 18°, 30°, 36°, 45°, 54°, 60° and 72° according to their 3D coordinates in the solids. Each of those angles is measured from the top most point of the circle when a vertical line is turned through that angle. Each line from the base point meets the circle at a point whose a height is 1 (72°), 1+Phi (60°), 2+Phi (54°), 2+2 Phi (45°), 2+3 Phi (36°), 3+3 Phi (30°) or 3+4 Phi (18°). Do look at his pages for more fascinating information on 120 3D solids, of which we will also explore the most symmetrical 5 on our next page.

Things to do 1. Suppose the origin of the circle is the lowest point and its radius is 2 + 2 Phi. Find the equation of the circle. 2. Use your answer to the previous question to find the coordinates of each of the points on the circle with the angles shown. 3. Compute the lengths of each of the red lines from the lowest point to the points shown on the circle. 4. From any two points A and B on a circle, the angle AOB at the centre of a circle, O, is twice the angle at any point on the circumference in the same sector. In the diagram,all the red angles at the circumference are equal; the red angles are twice the blue angle AOB at the centre; the red angles are to a point in the same sector of the circle as is the centre of the circle so they cannot be in the grey sector.

Use the above theorem to find three points on the circle ABOVE this Things To Do section where a line from the centre makes an angle with the vertical of i. 2×18=36° ii. 2×30=60° iii. 2×36=72°

Other angles with exact trig expressions involving square-roots Are there other angles with a simple exact expression for their cosine or sine? Well it all depends upon what you mean by simple! Carl Friedrich GAUSS (177 - 1855) looked at a similar problem which answers this question. He investigated if there was a method of constructing a regular polygon of n sides using only a pair of comes (to draw circles) and a straight-edge (a ruler with no markings). We know we can construct a regular polygon for all of the values of n=3, 4, 5, 6, 8 and 10. Halving There is a simple geometrical way to use comes to divide an angle into two (angle bisection). So all the angles in a regular n-gon can be split into two to make a regular 2n-gon. We can repeat the process to get a 4n-gon, 8n-gon and in general a 2kn-gon for any k once we have a method of constructing a regular n-gon. The Trig Formula section above contains a formula for the cosine of half an angle in of the cosine of the (whole) angle: c o s

A

1 + cos(A) =

2

2 + 2cos(A) =

2

2

As Mitch Wyatt pointed out to me, since we know that cos(90°) is 0 and 90° is /2 radians, we can use it to find the cosine of half that angle (45° or /4 radians) and then halve that angle again and so on. Each time we introduce another square root so we get a cascading or nested sequence of square roots: co s co

√ 2 4

= 2

s

√2√ 2 = +

8

2 co s

√ 2√2√ 2 + + =

16

2 co s

√

√ 2 2√ 2√ 2 = +++

32

2

However, this page is about sines and cosines which have simpler expressions, so we will not expand on this except to say that it shows how we can always find an exact expression for the sine (or cosine) of 1/2, 1/4, 1/8, ..., 1/2 , ... of any angle for which we have an exact sine (or cosine) expression. n

Superimposing If we construct a regular triangle (3 sides) and with the same circle centre, construct three regular pentagons (5 sides) with each having one vertex in common with the triangle, we will have the 15 vertices of a regular 15-gon. This is shown on the right with the 3 pentagons in blue on the same circle, each having a vertex in common with the red triangle and the regular 15-gon appears in yellow. By superimposing two regular polygons like this, we can construct a regular P×Q-gon (if P and Q have no factors in common otherwise more than one vertex of each will coincide). Do we know all the angles? All this was known in Euclid's time, around the year 300 BC. So what about 7ths and 9ths? Is it possible to find sines and cosines of all the multiples of 1/7 and 1/9 of a turn in exact (using square roots)? What about 11ths and 12ths etc.? In the next 2000 years no one found an exact geometric method for 7-gons or 9-gons but also no one had proved it was impossible to construct such regular polygons. Then C F Gauss completely solved the problem while he was a student at Göttingen between 1795 and 1798. Gauss found the conditions on n and its the prime factors to solve two equivalent problems: • drawing a regular n-sided polygon using only a straight edge and com and •

expressing the cos and sin of 360/n° using only square roots. If we factor n as 2ap1bp2c..., i.e. a, b, c, ... are the powers of n's prime factors: 2, p1, p2, ... (the prime's power is 0 if it is not a factor of n) then both of the problems are solvable when

•

b,c,... and all the powers except a, the power of 2, must be 1, and

•

the primes>2 that are factors of n (that is p1, p2, ...) must be of the form 22 +1 for some number k. k

Both problems are solvable for these values of n and only for these values. Prime numbers of the form 22 +1 are called Fermat primes. The series of numbers of the form 22 +1 begins k

k

22 + 1 = 3, 22 + 1 = 5, 22 + 1 = 17, 22 + 1 = 257, 22 + 1 = 65537, ... However not every number of the form 22 + 1 is prime -- and it is only the prime ones that we must have as factors of n. The next one, 22 + 1 is 4294967297 and has a factor of 641 so it is not prime. In fact, we do not know if there are any more primes of this form except the first 5 listed above. 0

1

2

3

4

k

5

Such numbers, n, of the form Gauss describes are as follows, one per row, each a product of some of the Fermat primes (but each prime at most once) followed by its multiples of two. For any number in the table, its double is also in the table: 24 36 5 10 3×5=15 30 17 34 3×17=51 102 5×17=75 150 3×5×17=225 450 257 514

8 12 20 60 68 204 300 900 1028

16 24 40 120 136 408 600 1800 2056

32 48 80 240 ... ... ... ... ...

... ... ... ...

When put in order, we have (1), 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, ... which is Sloane's A003401. Here, for instance, is the cosine of an angle involving 17ths:

8

co s

√1 √ 7 – 68 + 12√17 + 2(√17 – 1)S – 16 T 1 + =S –

17 wher√ e S 34 – 2√17 =

16 an √ d T 34 + 2√17 =

Things to do 1.

From the five 'starting values' above: 3, 5, 17, 257 and 65537 we can multiply these and also double any number to get a constructible polygon or an angle with a sine which involves nothing more than square-roots. Don't forget that we can double 2 to get 4 (a square), 8 (octagon), 16, etc too but we cannot use any of the 5 odd primes more than once in any product (so 3x3=9 and 5x5=25 are not in the list but 3x5x7 is). A complete list will involve these five numbers of course together with their products 3x5=15, 3x17=51, ... and you can double any of these any number of times: 2x3=6, 2x15=30 as well as 2x6=12, 2x12=24,... . a.

What are the first 12 values in the list that starts 3, 4, 5, ...?

b. Check that there are 24 values (excluding 1 and 2) less than 100. c. 2.

Is 100 in the list?

From the five known values: 3, 5, 17, 257 and 65537, there are a finite number of odd numbers n for which sin(360/n°) can be written with square-roots alone i.e. products of these 5 numbers where no number can be used more than once. How many odd numbers (excluding 1) can you make using these 5 no more than once in each product? This is the total number of known polygons we can construct with ruler and com or which have a sine (cosine) formed from nothing more than square-roots. With thanks to Richard Duffy for suggesting this puzzle

Tom Ace pointed out that there is more about this in chapter 15 of Oystein Ore's Number Theory and Its History from 1948 and now available as a Dover book(1988). If you are printing this page and the horizontal lines of fractions or the square-roots do not appear, make sure you have the "Print Images" and "Print Background" options checked when your browser's printing window appears. ll this was known in Euclid's time, around the year 300 BC. So what about 7ths and 9ths? Is it possible to find sines and cosines of all the multiples of 1/7 and 1/9 of a turn in exact (using square roots)? What about 11ths and 12ths etc.? In the next 2000 years no one found an exact geometric method for 7-gons or 9-gons but also no one had proved it was impossible to construct such regular polygons. Then C F Gauss completely solved the problem while he was a student at Göttingen between 1795 and 1798. Gauss found the conditions on n and its the prime factors to solve two equivalent problems: • drawing a regular n-sided polygon using only a straight edge and com and •

expressing the cos and sin of 360/n° using only square roots. If we factor n as 2ap1bp2c..., i.e. a, b, c, ... are the powers of n's prime factors: 2, p1, p2, ... (the prime's power is 0 if it is not a factor of n) then both of the problems are solvable when

•

b,c,... and all the powers except a, the power of 2, must be 1, and

•

the primes>2 that are factors of n (that is p1, p2, ...) must be of the form 22 +1 for some number k. k

Both problems are solvable for these values of n and only for these values. Prime numbers of the form 22 +1 are called Fermat primes. The series of numbers of the form 22 +1 begins k

k

22 + 1 = 3, 22 + 1 = 5, 22 + 1 = 17, 22 + 1 = 257, 22 + 1 = 65537, ... However not every number of the form 22 + 1 is prime -- and it is only the prime ones that we must have as factors of n. The next one, 22 + 1 is 4294967297 and has a factor of 641 so it is not prime. In fact, we do not know if there are any more primes of this form except the first 5 listed above. 0

1

2

3

4

k

5

Such numbers, n, of the form Gauss describes are as follows, one per row, each a product of some of the Fermat primes (but each prime at most once) followed by its multiples of two. For any number in the table, its double is also in the table: 24 36 5 10 3×5=15 30 17 34

8 12 20 60 68

16 24 40 120 136

32 48 80 240 ...

... ... ... ...

3×17=51 102 5×17=75 150 3×5×17=225 450 257 514

204 300 900 1028

408 600 1800 2056

... ... ... ...

When put in order, we have (1), 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, ... which is Sloane's A003401. Here, for instance, is the cosine of an angle involving 17ths:

8

co s

√1 √ 7 – 68 + 12√17 + 2(√17 – 1)S – 16 T 1 + =S –

17 wher√ e S 34 – 2√17 =

16 an √ d T 34 + 2√17 =

Things to do 1.

From the five 'starting values' above: 3, 5, 17, 257 and 65537 we can multiply these and also double any number to get a constructible polygon or an angle with a sine which involves nothing more than square-roots. Don't forget that we can double 2 to get 4 (a square), 8 (octagon), 16, etc too but we cannot use any of the 5 odd primes more than once in any product (so 3x3=9 and 5x5=25 are not in the list but 3x5x7 is). A complete list will involve these five numbers of course together with their products 3x5=15, 3x17=51, ... and you can double any of these any number of times: 2x3=6, 2x15=30 as well as 2x6=12, 2x12=24,... . a.

What are the first 12 values in the list that starts 3, 4, 5, ...?

b. Check that there are 24 values (excluding 1 and 2) less than 100. c. 2.

Is 100 in the list?

From the five known values: 3, 5, 17, 257 and 65537, there are a finite number of odd numbers n for which sin(360/n°) can be written with square-roots alone i.e. products of these 5 numbers where no number can be used more than once. How many odd numbers (excluding 1) can you make using these 5 no more than once in each product? This is the total number of known polygons we can construct with ruler and com or which have a sine (cosine) formed from nothing more than square-roots. With thanks to Richard Duffy for suggesting this puzzle

3.

Pythagorean Identities

4. 5.

6. Addition Formulas

7. 8.

9. Subtraction formulas

10. 11.

12.Double Angle Formulas

13. 14.

15.Half-Angle Formulas

16. 17.

18.Product Formulas

19. 20.

21.Factoring Formulas

23.The following two formulas are of only limited use:

22.

24.

Trigonometric Addition Formulas

Angle addition formulas express trigonometric functions of sums of angles

in of functions of

and . The fundamental formulas of angle addition in trigonometry are given by (1 ) (2 ) (3 ) (4 ) (5 )

(6 ) The first four of these are known as the prosthaphaeresis formulas, or sometimes as Simpson's formulas. The sine and cosine angle addition identities can be compactly summarized by the matrix equation (7 ) These formulas can be simply derived using complex exponentials and the Euler formula as follows. (8) (9) (10 ) (11 ) Equating real and imaginary parts then gives (1) and (3), and (2) and (4) follow immediately by substituting

for .

Taking the ratio of (1) and (3) gives the tangent angle addition formula (12 ) (13 ) (14 ) (15 ) The double-angle formulas are (16 ) (17 ) (18 ) (19 )

(20 ) Multiple-angle formulas are given by (21 ) (22 ) and can also be written using the recurrence relations (23 ) (24 ) (25 )

The angle addition formulas can also be derived purely algebraically without the use of complex numbers. Consider the small right triangle in the figure above, which gives (26 ) (27 ) Now, the usual trigonometric definitions applied to the large right triangle give (28 ) (29 )

(30 ) (31 ) Solving these two equations simultaneously for the variables

and

then immediately gives (32 ) (33 )

These can be put into the familiar forms with the aid of the trigonometric identities (34 ) and (35 ) (36 ) (37 ) (38 ) which can be verified by direct multiplication. Plugging (◇) into (◇) and (38) into (◇) then gives (39 ) (40 ) as before.

A similar proof due to Smiley and Smiley uses the left figure above to obtain (41 ) from which it follows that (42 ) Similarly, from the right figure, (43 ) so (44 )

Similar diagrams can be used to prove the angle subtraction formulas (Smiley 1999, Smiley and Smiley). In the figure at left, (45 ) (46 ) (47 ) giving (48 ) Similarly, in the figure at right, (49 ) (50 )

(51 ) giving (52 )

A more complex diagram can be used to obtain a proof from the

identity (Ren 1999). In the above figure, let

. Then (53 )

An interesting identity relating the sum and difference tangent formulas is given by (54 ) (55 )