a  cos(a) sin(b) 
tan(a) cot(b) 
b  

radians  degrees  degrees  radians  
0  0  1  0  90  π 2 
π 24 
7.5  `(sqrt(2+sqrt(2+sqrt 3)))/2`  `sqrt 6  sqrt 3 + sqrt 2  2` `= (sqrt 2  1)(sqrt 3  sqrt 2)` 
82.5  11 π 24 
π 12 
15 
`(sqrt 6 + sqrt 2)/4 = sqrt(1/2+(sqrt 3)/4) = sqrt((4+ sqrt 12)/8)` 
`2sqrt 3` 
75  5 π 12 
cos^{2}(15°)=[0;1,13, 1,12]  tan(15°) = [0; 3, 1, 2 ] tan^{2}(15°) = [0; 13, 1, 12]  
π 10 
18 
`sqrt(10+2 sqrt 5)/4 = sqrt((5+sqrt 5)/8)` 
`sqrt(1 (2 sqrt 5)/5)`  72  2 π 5 
cos^{2}(18°) = [0; 1, 9, 2, 8]  tan^{2}(18°) = [0; 9, 2,8]  
π 8 
22.5  `(sqrt(2+sqrt 2))/2 = sqrt((4+sqrt 8)/8)`  `sqrt 2  1`  67.5  3 π 8 
cos^{2}(22.5°) = [0; 1, 5, 1, 4] 
tan(22.5°) = [0; 2] tan^{2}(22.5°) = [0; 5, 1, 4]  
π 6 
30  ` (sqrt 3)/2`  `(sqrt 3)/3`  60  π 3 
cos(30°) = [ 0; 1, 6, 2, 6 ] cos^{2}(30°) = [0; 1, 3] 
tan(30°) = [ 0; 1, 1, 2 ] tan^{2}(30°) = [0; 3 ]  
π 5 
36  `(sqrt 5 + 1)/4 = sqrt((3+sqrt 5)/8)`  `sqrt(5  2 sqrt 5)`  54  3 π 10 
cos(36°) = [ 0; 1, 4 ] cos^{2}(36°) = [ 0; 1, 1, 1, 8, 2 ]  tan^{2}(36°) = [ 0; 1, 1, 8, 2 ]  
5 π 24 
37.5  `sqrt(2 + sqrt( 2  sqrt 3))/2 `  `sqrt 6 + sqrt 3  sqrt 2  2` `= (sqrt 2 + 1)(sqrt 3  sqrt 2)` 
52.5  7 π 24 
π 4 
45  `sqrt 2/2`  `1`  45  π 4 
cos(45°) = [ 0; 1, 2 ] cos^{2}(45°) = [ 0; 2 ]  
7 π 24 
52.5  `(sqrt(2  sqrt( 2  sqrt 3)))/2` 
`sqrt 6  sqrt 3  sqrt 2 + 2` `= (sqrt 2  1)(sqrt 3 + sqrt 2)` 
37.5  5 π 24 
3 π 10 
54  `(sqrt(10  2 sqrt 5))/4 = sqrt((5  sqrt 5)/8)`  `sqrt((5 + 2 sqrt 5)/5)`  36  π 5 
cos^{2}(54°) = [ 0; 2, 1, 8, 2 ]  tan^{2}(54°) = [ 1; 1, 8, 2 ]  
π 3 
60  `1/2`  `sqrt 3`  30  π 6 
cos(60°) = [0; 2] cos^{2}(60°) = [0; 4] 
tan(60°) = [ 1; 1, 2 ] tan^{2}(60°) = 3  
3 π 8 
67.5  `sqrt(2sqrt 2)/2 = sqrt((4sqrt 8)/8)`  `1 + sqrt 2`  22.5  π 8 
cos^{2}(67.5°) = [ 0; 6, 1, 4 ] 
tan(67.5°) = [ 2; 2 ] tan^{2}(67.5°) = [ 5; 1, 4 ]  
2 π 5 
72  `(sqrt 5  1)/4 = sqrt((3sqrt 5)/8)`  `sqrt(5 + 2 sqrt 5)`  18  π 10 
cos(72°) = [ 0; 3, 4 ] cos^{2}(72°) = [ 0; 10, 2, 8 ]  tan^{2}(72°) = [ 9; 2, 8 ]  
5 π 12 
75  `(sqrt 6  sqrt 2)/4 = sqrt(1/2  sqrt 3/4) = sqrt((4  sqrt 12)/8)`  `2+sqrt 3`  15  π 12 
cos^{2}(75°) = [ 0; 14, 1, 12 ] 
tan(75°) = [ 3; 1, 2 ] tan^{2}(75°) = [ 13; 1, 12 ]  
11 π 24 
82.5  `sqrt(2  sqrt(2 + sqrt 3))/2`  `sqrt 6 + sqrt 3 + sqrt 2 + 2` `= (sqrt 2 + 1)(sqrt 3 + sqrt 2)` 
7.5  π 24 
π 2 
90  `0`  `oo`  0  0 
If you have Mathematica's (free)
CDF player installed (get it here)
then you can click anywhere to change the angle.
How the Trigonometry Functions Are Related from the Wolfram Demonstrations Project by C. Ormullion

Here is a static image which is a useful diagram anyway: 
Hoever, the diagram is useful for seeing what is measured by the six functions
but it does not give any indication of the signs of the values.
The signs are chosen to make the trig formula see below consistent for all angles. Also,
the next section shows their graphs and from the basic SINE graph, all the others follow.
The sine function
has many applications in mechanics (e.g. the motions of rotating objects),
electronics (e.g. alternating electric current and electromagnetic waves), ... .
Angle ° rad 
cosine sine 
Angle ° rad  

0  0 
 90  π 2  
30  π 6 
 60  π 3  
45  π 4 
 45  π 4  
60  π 3 
 30  π 6  
90  π 2 
 0  0 

Angle°  cosine sine  Angle° 

72  $$\frac{\sqrt{2  \Phi}}{2}$$  18 
54  $$\frac{\sqrt{2  \varphi}}{2}$$  36 
36  $$\frac{\sqrt{2 + \varphi}}{2}$$  54 
18  $$\frac{\sqrt{2 + \Phi}}{2}$$  72 
cos(9°= π/20) =  \(\frac{1}{2} \sqrt{2 + \sqrt{2 + \Phi}} \)  = sin(81°) 
cos(18°=π/10) =  \(\frac{1}{2} \sqrt{2 + \sqrt{2 + \varphi}} = \frac{1}{2}\sqrt{2 + \Phi} \)  = sin(72°) 
cos(27°=3π/20) =  \(\frac{1}{2} \sqrt{2 + \sqrt{2  \varphi}}\)  = sin(63°) 
cos(36°=π/5) =  \(\frac{1}{2} \sqrt{2 + \sqrt{2  \Phi}} = \frac{1}{2}\sqrt{2 + \varphi} = \frac{\Phi}{2}\)  = sin(54°) 
cos(45°=π/4) =  \(\frac{1}{2} \sqrt{2 \pm \sqrt{2  2}} = \frac{\sqrt{2}}{2}\)  = sin(45°) 
cos(54°=3π/10) =  \(\frac{1}{2} \sqrt{2  \sqrt{2  \Phi}} = \frac{1}{2}\sqrt{2  \varphi}\)  = sin(36°) 
cos(63°=7π/20) =  \(\frac{1}{2} \sqrt{2  \sqrt{2  \varphi}}\)  = sin(27°) 
cos(72°=2π/5) =  \(\frac{1}{2} \sqrt{2  \sqrt{2 + \varphi}} = \frac{1}{2}\sqrt{2  \Phi} = \frac{\varphi}{2}\)  = sin(18°) 
cos(81°=9π/20) =  \(\frac{1}{2} \sqrt{2  \sqrt{2 + \Phi}}\)  = sin(9°) 
cos(0)  =  \(\frac{1}{2} \sqrt{2 + \sqrt{2 + 2}} = 1\)  sin(90°) 
cos(7.5°=π/24)  =  \(\frac{1}{2} \sqrt{2 + \sqrt{2 + \sqrt{3}}}\)  sin(82.5°) 
cos(37.5°=5π/24)  =  \(\frac{1}{2} \sqrt{2 + \sqrt{2  \sqrt{3}}}\)  sin(52.5°) 
cos(45°=π/4)  =  \(\frac{1}{2} \sqrt{2 \pm \sqrt{2  2}} = \frac{\sqrt{2}}{2}\)  sin(45°) 
cos(52.5°=7π/24)  =  \(\frac{1}{2} \sqrt{2  \sqrt{2  \sqrt{3}}}\)  sin(37.5°) 
cos(82.5°=11π/24)  =  \(\frac{1}{2} \sqrt{2  \sqrt{2 + \sqrt{3}}}\)  sin(7.5°) 
cos(90°=π/2)  =  \(\frac{1}{2} \sqrt{2  \sqrt{2 + 2}} = 0\)  sin(0°) 
This triangle is just a square cut along a diagonal. If the sides are of length 1, the diagonal is length √2. This gives the sin, cos and tan of 45°.  Here is an equilateral triangle where all sides and all angles are equal (to 60°). If the sides are of length 2, 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°. 
Phi = = 1.618033988.. =  1 + √5  = 1 +  1 
2  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°. 
a^{2}  = b^{2} + c^{2} – 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:  
a^{2}  = 2^{2} + 2^{2} – 2 x 2 x 2 x cos(30°)  
= 8 – 4 √3  
= 2 (4 – 2 √3)  
But (√3 – 1)^{2} = 3 + 1 – 2 √3 = 4 – 2 √3 and so  
a^{2}  = 2 (√3 – 1)^{2}  
Taking the squareroot:  
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 rightangled 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°: 
sin(x) = a / h cos(x) = b / h tan(x) = a / b cot(x) = b / a 

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)  tan(A + B) = 
 
sin( A – B )  = sin(A)cos(B) – cos(A)sin(B)  
cos( A + B )  = cos(A)cos(B) – sin(A)sin(B)  tan(A – B) = 
 
cos( A – B )  = cos(A)cos(B) + sin(A)sin(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 sin^{2}(A) cos( 2A ) = cos^{2}(A) – sin^{2}(A) cos( 2A ) = 2 cos^{2}(A) – 1 
 


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.
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.
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 compasses
(to draw circles) and a straightedge
(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.
The Trig Formula section above contains a formula for the cosine of half an angle in terms of the cosine of the (whole) angle:
cos 
 = 
 = 

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}/_{2n}, ... of any angle for which we have an exact sine (or cosine) expression.
cos π
4=
√ 2
2
cos π
8=
√
2 +
√ 2
2
cos π
16=
√
2 +
√
2 +
√ 2
2
cos π
32=
√
2 +
√
2 +
√
2 +
√ 2
2
In the next 2000 years no one found an exact geometric method for 7gons or 9gons 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 prime factors to solve two equivalent problems:
Prime numbers of the form 2^{2k}+1 are called Fermat primes. The series of numbers of the form 2^{2k}+1 begins
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:
2  4  8  16  32  64  128  256  ...  
3  6  12  24  48  96  192  ...  
5  10  20  40  80  160  ...  
3×5=  15  30  60  120  240  ...  
17  34  68  136  272  ...  
3×17=  51  102  204  408  ...  
5×17=  85  170  340  680  ...  
3×5×17=  255  510  1020  2040  ...  
257  514  1028  2056  ... 
The odd terms (the left hand column apart from 2) is the series
Some interesting facts about these odd numbers (we include 1 here too) are:
1 =  1_{2} 
3 =  11_{2} 
5 =  101_{2} 
15 =  1111_{2} 
17 =  10001_{2} 
85 =  110011_{2} 
... 


A_{+} = 15 + √17  A_{–} = 15 – √17;  
B_{+} = √2 √  B_{–} = √2 √  
C_{+} = 17 + 3 √17  C_{–} = 17 – 3 √17;  
D_{+} = √2 √  D_{–} = √2 √ 
a  32 cos^{2}(a) 

π 17 
A_{+} + B_{–} + 2 √C_{+} – D_{+} 
2 π 17 
A_{+} – B_{–} + 2 √C_{+} + D_{+} 
3 π 17 
A_{–} + B_{+} + 2 √C_{–} + D_{–} 
4 π 17 
A_{+} + B_{–} – 2 √C_{+} – D_{+} 
5 π 17 
A_{–} + B_{+} – 2 √C_{–} + D_{–} 
6 π 17 
A_{–} – B_{+} + 2 √C_{–} – D_{–} 
7 π 17 
A_{–} – B_{+} – 2 √C_{–} – D_{–} 
8 π 17 
A_{+} – B_{–} – 2 √C_{+} + D_{+} 
The next odd value is `pi/257` and the exact expression for `cos(pi/257)` using squareroots only will be even more complicated!
© 19962013 Dr Ron Knott
updated 2 September 2013 