# Exact trigonometric constants

Exact constant expressions for trigonometric expressions are sometimes useful, mainly for simplifying solutions into radical forms which allow further simplification.

All values of sine, cosine, and tangent of angles with 3° increments are derivable using identities: Half-angle, Double-angle, Addition/subtraction and values for 0°, 30°, 36° and 45°. Note that 1° = π/180 radians.

 Contents

## Table of constants

Values outside 0° ... 45° angle range are trivially extracted from circle axis reflection symmetry from these values.

### 0° Fundamental

[itex]\sin 0^\circ = 0 [itex]
[itex]\cos 0^\circ = 1 [itex]
[itex]\tan 0^\circ = 0 [itex]

### 3° - 60-sided polygon

sin(3°) = [2√(5 + √5)(1 − √3) + √2(√5 − 1)(√3 + 1)]/16
cos(3°) = [2√(5 + √5)(1 + √3) + √2(√5 − 1)(√3 − 1)]/16
tan(3°) = [(2 − √3)(3 + √5) − 2](2 − √(2(5 − √5)))/4

### 6° - 30-sided polygon

sin(6°) = [√(6(5 − √5)) − (√5 + 1)]/8
cos(6°) = [√(2(5 − √5)) + √3(√5 − 1)]/8
tan(6°) = [√(5 − 2√5)(√5 + 1) + √3(1 − √5)]/2

### 9° - 20-sided polygon

sin(9°) = [−2√(5 − √5) + √2(√5 + 1)]/8
cos(9°) = [+2√(5 − √5) + √2(√5 + 1)]/8
tan(9°) = −√(5 − 2√5)(2 + √5) + (√5 + 1)

### 12° - 15-sided polygon

[itex]\sin \frac{\pi}{15} = \sin 12^\circ = \frac{\sqrt{2(5+\sqrt5)}-\sqrt 3 (\sqrt 5 -1)}{8} [itex]
[itex]\cos \frac{\pi}{15} = \cos 12^\circ = \frac{\sqrt{6(5+\sqrt5)}+(\sqrt 5 -1)}{8} [itex]
[itex]\tan(12^\circ)=(\sqrt{5-2\sqrt{5}}\,(2+\sqrt{5})+(\sqrt{5}\,+1)/2[itex]

### 15° - 12-sided polygon

[itex]\sin \frac{\pi}{12} = \sin 15^\circ = \frac{\sqrt 2 \cdot \left(\sqrt 3 - 1\right)}{4}[itex]
[itex]\cos \frac{\pi}{12} = \cos 15^\circ = \frac{\sqrt 2 \cdot \left(\sqrt 3 + 1\right)}{4}[itex]
[itex]\tan \frac{\pi}{12} = \tan 15^\circ = 2 - \sqrt 3[itex]
[itex]\cot \frac{\pi}{12} = \cot 15^\circ = 2 + \sqrt 3[itex]

### 18° - 10-sided polygon

[itex]\sin \frac{\pi}{10} = \sin 18^\circ = \frac{\sqrt 5 - 1}{4}[itex]
[itex]\cos \frac{\pi}{10} = \cos 18^\circ = \frac{\sqrt{2(5 + \sqrt 5)}}{4} [itex]
[itex]\tan \frac{\pi}{10} = \tan 18^\circ = \frac{\sqrt{5(5 - 2 \sqrt 5)}}{5} [itex]
[itex]\cot \frac{\pi}{10} = \cot 18^\circ = \sqrt{5 + 2 \sqrt 5} [itex]

### 21° - Sum 9° + 12°

sin(21°) = [2√(5 − √5)(√3 + 1) − √2(√3 − 1)(1 + √5)]/16
cos(21°) = [2√(5 − √5)(√3 − 1)+√2(√3 + 1)(1 + √5)]/16
tan(21°) = [√(5 − 2√5)(1 + 2√3 − √5) + (2 + √3)(√5 − 3) + 2]/2

### 22.5° - Octagon

[itex]\begin{matrix}

\sin \frac{\pi}{8} & = & \sin 22.5^\circ & = & \frac{\sqrt{2 - \sqrt{2}}}{2} \\ \cos \frac{\pi}{8} & = & \cos 22.5^\circ & = & \frac{\sqrt{2 + \sqrt{2}}}{2} \\ \tan \frac{\pi}{8} & = & \tan 22.5^\circ & = & \sqrt{2}-1 \\ \cot \frac{\pi}{8} & = & \cot 22.5^\circ & = & \sqrt{2}+1 \\ \end{matrix}[itex]

### 24° - Sum 12° + 12°

sin(24°) = √(2(5+√5))(1-√5)+2√3(1+√5))/16
cos(24°) = √(6(5+√5))(√5-1)+2(1+√5))/16
tan(24°) = (√(10+2√5)-2√3)(3+√5)/4
cotan(24°) = (√(10+2√5)+2√3)(√5-1)/4

### 27° - Sum 12° + 15°

sin(27°) = ((2√(5+√5)+√2(1-√5))/8
cos(27°) = ((2√(5+√5)+√2(√5-1))/8
tan(27°) = -√(5-2(√5))+(√5-1)

### 30° - Hexagon

[itex]\sin \frac{\pi}{6} = \sin 30^\circ = \frac{1}{2}[itex]
[itex]\cos \frac{\pi}{6} = \cos 30^\circ = \frac{\sqrt 3}{2}[itex]
[itex]\tan \frac{\pi}{6} = \tan 30^\circ = \frac{\sqrt 3}{3}[itex]
[itex]\cot \frac{\pi}{6} = \cot 30^\circ = \frac{3}{\sqrt 3} = \sqrt 3[itex]

### 33° - Sum 15° + 18°

sin(33°) = (2√(5+√5)(-1+√3)+√2(√5-1)(1+√3))/16
cos(33°) = (2√(5+√5)(+1+√3)+√2(√5-1)(1-√3))/16
tan(33°) = (√(5(5-2√5))(-15+10√3-7√5+4√15)+5((-2+√3)(3+√5)+2))/10

### 36° - Pentagon

[itex]\sin \frac{\pi}{5} = \sin 36^\circ = \frac{\sqrt{2(5 - \sqrt 5)} }{4}[itex]
[itex]\cos \frac{\pi}{5} = \cos 36^\circ = \frac{\sqrt 5+1}{4}[itex]
[itex]\tan \frac{\pi}{5} = \tan 36^\circ = \sqrt{5 - 2\sqrt 5} [itex]

### 39° - Sum 18°+ 21°

sin(39°) = (2√(5-√5)(1-√3)+√2(+1+√3)(1+√5))/16
cos(39°) = (2√(5-√5)(1+√3)+√2(-1+√3)(1+√5))/16
tan(39°) = (√(2(5+√5))-2)((2-√3)(-3+√5)+2)/4

### 42° - Sum 21° + 21°

sin(42°) = (√(6(5-√5))(1+√5)+2(1-√5))/16
cos(42°) = (√(2(5-√5))(1+√5)+2√3(-1+√5))/16
tan(42°) = (-√(5-2√5)(3+√5)+√3(1+√5))/2

### 45° - Square

[itex]\sin \frac{\pi}{4} = \sin 45^\circ = \frac{\sqrt 2}{2}[itex]
[itex]\cos \frac{\pi}{4} = \cos 45^\circ = \frac{\sqrt 2}{2}[itex]
[itex]\tan \frac{\pi}{4} = \tan 45^\circ = 1[itex]
[itex]\cot \frac{\pi}{4} = \cot 45^\circ = 1[itex]

## Notes

### Uses for constants

As an example of the use of these constants, the volume of a dodecahedron is

V = 5e3cos(36°)/tan2(36°)

Using

cos(36°) = (√5 + 1)/4
tan(36°) = √(5 − 2√5)

this can be simplified to:

V = e3(15 + 7√5)/4.

### Derivation triangles

Missing image
Polygontriangle.gif
Regular polygon (N-sided) and its fundamental right triangle. Angle: a = 180/N °

The derivation of sine, cosine, and tangent constants into radial forms is based upon the constructability of right triangles.

Here are right triangles made from symmetry sections of regular polygons are used to calculate fundamental trigonometric ratios. Each right triangle represents three points in a regular polygon: a vertex, an edge center containing that vertex, and the polygon center. A N-agon can be divided into 2N right triangle with angles of {180/N, 90−180/N, 90} degrees, for N = 3, 4, 5, ...

Constructibility of 3, 4, 5, and 15 sided polygons are the basis, and angle bisectors allow multiples of two to also be derived.

• Constructable
• 3×2X-sided regular polygons, X = 0, 1, 2, 3, ...
• 4×2X-sided
• 45°-45°-90° triangle - square (4-sided)
• 67.5°-22.5°-90° triangle - octagon (8-sided)
• 88.75°-11.25°-90° triangle - hexakaidecagon (16-sided)
• ...
• 5×2X-sided
• 54°-36°-90° triangle - pentagon (5-sided)
• 72°-18°-90° triangle - decagon (10-sided)
• 81°-9°-90° triangle - icosagon (20-sided)
• 85.5°-4.5°-90° triangle - tetracontagon (40-sided)
• 87.75°-2.25°-90° triangle - octacontagon (80-sided)
• ...
• 15×2X-sided
• ... (Higher constructible regular polygons don't make whole degree angles: 17, 51, 85, 255, 257...)

### Expressions not unique

Simplifying nested radical expressions is nontrivial. The expressions here may not all be fully reduced.

Example:

[itex]
 4\sin 18^\circ
= \sqrt{2(3 - \sqrt 5)}
= \sqrt 5 - 1


[itex]

It's not evident that this simplification is equivalent, and in general nested radicals can not be reduced.

In general this is reducible:
[itex]
 \sqrt{a + b\sqrt c} = d + e\sqrt c


[itex], if a2 − 4b2c is a perfect square.

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