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.

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

that are expressible as simple terms involving square-roots.

a cos(a)
sin(b)
tan(a)
cot(b)
b
0 01 0 90
 2
 12
15
 6 + 2 4 4
=
1 +
 3

2

4
=
4 +
 12

8

8
cos2(15°) = [0; 1, 13, [1, 12]]
2 – 3
tan(15°) = [0; 3, [1, 2]]
tan2(15°) = [0; 13, [1, 12]]
75
 5 12
 10
18
10 + 2
 5
=
5 +
 5

8

8

4
cos2(18°) = [0; 1, 9, [2, 8]]
 1 – 25 5

tan2(18°) = [0; 9, [2,8]]
72
 2 5
 8
22·5
 2 + 2

2
=
4 +
 8

8

8
cos2(22·5°) = [0; 1, 5, [1, 4]]
2 – 1
tan(22·5) = [0; [2]]
tan2(22·5) = [0; 5, [1,4]]
67·5
 3 8
 6
30
 3 4
=
 3 2
cos(30°) = [0; 1, 6, [2, 6]]
cos2(30°) = [0; 1, 3]
 3 3

tan(30°) = [0; 1, [1,2]]
tan2(30°) = [0; 3]
60
 3
 5
36
 5 + 1 4 4
=
 3 + 5 8 8
cos(36°) = [0; 1, [4]]
cos2(36°) = [0; 1, 1, 1, [8, 2]]
 5 – 25
tan2(36°) = [0; 1, 1, [8, 2]]
54
 3 10
 4
45
 2 4
=
 1 2
cos(45°) = [0; 1, [2]]
cos2(45°) = [0; 2]
1 45
 4
 3 10
54
10 – 2
 5
=
5 –
 5

8

8

4
cos2(54°) = [0; 2, 1, [8, 2]]
 1 + 25 5

tan2(54°) = [1; 1, [8, 2]]
36
 5
 3
60
 1 4
=
 1 2
cos(60°) = [0; 2]
cos2(60°) = [0; 4]
3
tan(60°) = [1; [1,2]]
tan2(60°) = 3
30
 6
 3 8
67·5
 2 – 2

2
=
4 –
 8

8

8
cos2(67·5) = [0; 6, [1, 4]]
1 + 2
tan(67·5) = [2; [2]]
tan2(67·5) = [5; [1,4]]
22·5
 8
 2 5
72
 5 – 1 4 4
=
 3 – 5 8 8
cos(72°) = [0; 3, [4]]
cos2(72°) = [0; 10, [2, 8] ]
 5 + 25

tan2(72°) = [9; [2, 8]]
18
 10
 5 12
75
 6 – 2 4 4
=
1 –
 3

2

4
=
4 –
 12

8

8
cos2(75°) = [0; 14, [1, 12]]
2 + 3
tan(75°) = [3; [1, 2]]
tan2(75°) = [13; [1,12]]
15
 12
 2
90 0 infinity 0 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.

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

b +
1

c +
 1 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°

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: sincostan (°)

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:

# Patterns

## The Simple Square-Root pattern

Ernesto La Orden of Madrid pointed out the following neat way to connect and remember the easiest of the sines (cosines):
Angle sine
cosine
Angle
90
 √4 =  1 2
0
60
 √3 2
30
45
 √2 = 1 2 √2
45
30
 √1 = 1 2 2
60
0
 √0 = 0 2
90

## The √(2 ± Phi) pattern

cos(9)  =
 1 2
2 +
 2 +
cos(18)  =
 1 2
2 +
 2 +
=
 1 2
 2 +
cos(27)  =
 1 2
2 +
 2 –
cos(36)  =
 1 2
2 +
 2 –
=
 1 2
 2 +
=
 2
cos(54)  =
 1 2
2 –
 2 –
=
 1 2
 2 –
cos(63)  =
 1 2
2 –
 2 –
cos(72)  =
 1 2
2 –
 2 +
=
 1 2
 2 –
=
 2
cos(81)  =
 1 2
2 –
 2 +
This pattern uses the identities
=
 2 –
and     =
 2 +
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:
Anglecosine
sine
Angle
90
 2 – √4
=  0
2
0
75
 2 – √3
2
15
67.5
 2 – √2
2
22.5
60
 2 – √1
=  1
22
30
45
 2 – √0
=  1
2√2
45
30
 2 + √1
=  √3
22
60
22.5
 2 + √2
2
67.5
15
 2 + √3
2
15
0
 2 + √4
=  1
2
0
Anglecosine
sine
Angle
72
 2 – Phi
2
18
54
 2 – phi
2
36
36
 2 + phi
2
54
18
 2 + Phi
2
72
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 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°.

### 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:
 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°.

### 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)2 = 3 + 1 – 2 3 = 4 – 2 3 and so 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 remember 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) = a / h
cos(x) = b / h
tan(x) = a / b
cot(x) = b / a
sin(x)  = cos(90° – x)   sin2(x) + cos2(x) = 1
 tan(x) = sin(x) cos(x)

 cot(x) = 1 = cos(x) tan(x) sin(x)

 tan2(x) + 1 = 1 cos2(x)

 cot2(x) + 1 = 1 sin2(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)

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

sinA =
 1 – cos(A) 2

2

cosA =
 1 + cos(A) 2

2

## 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. The angle at the centre of a circle is twice the angle at the circumference from the same chord (a line joining two points on the circle)
Use the above theorem to find three points on the circle where a line from the centre makes an angle with the vertical of
1. 2×18=36°
2. 2×30=60°
3. 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 compasses (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 compasses 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 terms of the cosine of the (whole) angle:

cos
 A 2
=
 1 + cos(A) 2
=
 2 + 2cos(A)

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 /8 radians) and then halve that angle and so on. Each time we introduce another square-root to we get a cascading or nested sequence of increasingly complex expressions:
 cos 4
=
 2

2
 cos 8
=
2 +
 2

2
 cos 16
=
2 +
2 +
 2

2
 cos 32
=
2 +
2 +
2 +
 2

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 half of any angle for which we have an exact sine (or cosine) expression.

### 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 terms (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 Gauss completely solved the problem while he was a student at Göttingen between 1795 and 1798. Gauss found the conditions on the primes and their powers to solve two equivalent problems:

• drawing a regular n-sided polygon using only a straight edge and compass 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, ... 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 22q+1 for some number q. Such primes are called Fermat primes. The series of the allowable primes after 2 is
220 + 1 = 3, 221 + 1 = 5, 222 + 1 = 17, 223 + 1 = 257, 224 + 1 = 65537
The next one, 225 + 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 first 5 listed above.
Both problems are solvable only for these values of n and only for these values.

For instance:

cos8  =
17 – 1 + S –
 68 + 1217 + 2(17 – 1)S – 16 T

17

16

where S =
 34 – 217
and T =
 34 + 217

#### 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,... .
1. What are the first 12 values in the list that starts 3, 4, 5, ...?
2. Check that there are 24 values (excluding 1 and 2) less than 100.
3. 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?
This is the total number of known polygons we can construct with ruler and compass 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 but now available as a Dover book (1988).
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