# Table of exact trigonometric functions

2011-10-19

Since I noticed I had to keep looking some of these up, I place them here, just for reference, gathered from around the internet. Many of these formulas can be written in different ways, but I have simplified them as much as possible.

Where, $\phi=\frac{1+\sqrt{5}}{2}=1.6180339887498948482045868\ldots$, also known as the golden ratio (phi).

a sin(a) = cos(b) = sin(90-b) = cos(90-a) tan(a) = cot(b) = tan(90-b) = cot(90-a) b
0 0 0 0 90 π/2
3 π/60 $\frac{1}{16}\left[(2-2\sqrt3)\sqrt{5+\sqrt5}+\sqrt2(-1+\sqrt{5})(1+\sqrt3)\right]$ $\frac14\left[(2-\sqrt{3})(3+\sqrt{5})-2\right]\left[2-\sqrt{10-2\sqrt{5}}\right]$ 87 29π/30
6 π/30 $\frac{1}{8} \left[\sqrt{30-6\sqrt5}-\sqrt5-1\right]$ $\frac12\left[\sqrt{10-2\sqrt5}-\sqrt3(-1+\sqrt5)\right]$ 84 7π/15
9 π/20 $\frac12\sqrt{2-\sqrt{2+\phi}}=\frac12\sqrt{2-\sqrt{\frac{5+\sqrt5}{2}}}$ $1+\sqrt5-\sqrt{5+2\sqrt5}$ 81 9π/20
12 π/15 $\frac{1}{8}\left[\sqrt{10+2\sqrt5}-\sqrt3(-1+\sqrt5)\right]$ $\frac12\left[\sqrt3(3-\sqrt5)-\sqrt{2(25-11\sqrt5)}\right]$ 78 13π/30
15 π/12 $\frac{\sqrt{2-\sqrt{3}}}{2}$ $2-\sqrt{3}$ 75 5π/12
18 π/10 $\frac{1}{2\phi}=\frac{1}{1+\sqrt{5}}=\frac{\sqrt{5}-1}{4}$ $\frac15\sqrt{25-10\sqrt5}$ 72 2π/5
21 7π/60 $\frac{1}{16}\left[2(\sqrt3+1)\sqrt{5-\sqrt5}-\sqrt2(\sqrt3-1)(1+\sqrt5)\right]$ $\frac{1}{4}\left[2-(2+\sqrt3)(3-\sqrt5)\right]\left[2-\sqrt{2(5+\sqrt5)}\right]$ 69 23π/60
22.5 π/8 $\frac{1}{2} \sqrt{2 – \sqrt{2}}$ $\sqrt{2} – 1$ 67.5 3π/8
24 2π/15 $\frac{1}{8}\left[-\sqrt{10-2\sqrt5}+\sqrt3(1+\sqrt5)\right]$ $\frac12\left[-\sqrt3(3+\sqrt5)+\sqrt{2(25+11\sqrt5)}\right]$ 66 11π/30
27 3π/20 $\frac12\sqrt{2-\sqrt{2-\frac{1}{\phi}}}=\frac12\sqrt{2-\sqrt{\frac{5-\sqrt5}{2}}}$ $-1+\sqrt5-\sqrt{5-2\sqrt5}$ 63 7π/20
30 π/6 $\frac12$ $\frac{1}{\sqrt{3}}$ 60 π/3
33 11π/60 $\frac{1}{16}\left[2(\sqrt3-1)\sqrt{5+\sqrt5}+\sqrt2(1+\sqrt3)(\sqrt5-1)\right]$ $\frac{1}{4}\left[2-(2-\sqrt3)(3+\sqrt5)\right]\left[2+\sqrt{2(5-\sqrt5)}\right]$ 57 19π/60
36 π/5 $\frac12\sqrt{2-\frac{1}{\phi}}=\frac12\sqrt{\frac{\sqrt5}{\phi}}=\sqrt{\frac{5-\sqrt5}{8}}$ $\sqrt{5-2\sqrt5}$ 54 3π/10
39 13π/60 $\frac1{16}[2(1-\sqrt3)\sqrt{5-\sqrt5}+\sqrt2(\sqrt3+1)(\sqrt5+1)]$ $\frac14\left[(2-\sqrt3)(3-\sqrt5)-2\right]\left[2-\sqrt{2(5+\sqrt5)}\right]$ 51 17π/60
42 7π/30 $\frac{1}{8} \left[\sqrt{30+6\sqrt5}-\sqrt5+1\right]$ $\frac12\left[-\sqrt{10+2\sqrt5}+\sqrt3(1+\sqrt5)\right]$ 48 4π/15
45 π/4 $\frac{1}{\sqrt{2}}$ 1 45 π/4
48 4π/15 $\frac{1}{8}\left[\sqrt{10+2\sqrt5}+\sqrt3(-1+\sqrt5)\right]$ $\frac12\left[\sqrt3(3-\sqrt5)+\sqrt{2(25-11\sqrt5)}\right]$ 42 7π/30
51 17π/60 $\frac1{16}[2(1+\sqrt3)\sqrt{5-\sqrt5}+\sqrt2(\sqrt3-1)(\sqrt5+1)]$ $\frac14\left[(2+\sqrt3)(3-\sqrt5)-2\right]\left[2+\sqrt{10+2\sqrt5}\right]$ 39 13π/60
54 3π/10 $\frac{\phi}{2}=\frac{1+\sqrt5}{4}$ $\frac15\sqrt{25+10\sqrt5}$ 36 π/5
57 19π/60 $\frac{1}{16}\left[2(\sqrt3+1)\sqrt{5+\sqrt5}+\sqrt2(1-\sqrt3)(\sqrt5-1)\right]$ $\frac{1}{4}\left[2-(2+\sqrt3)(3+\sqrt5)\right]\left[2-\sqrt{2(5-\sqrt5)}\right]$ 33 11π/60
60 π/3 $\frac{\sqrt{3}}{2}$ $\sqrt{3}$ 30 π/6
63 7π/20 $\frac12 \sqrt{2+\sqrt{2-\frac{1}{\phi}}}=\frac12\sqrt{2+\sqrt{\frac{5-\sqrt5}{2}}}$ $-1+\sqrt5+\sqrt{5-2\sqrt5}$ 27 3π/20
66 11π/30 $\frac{1}{8} \left[\sqrt{30-6\sqrt5}+\sqrt5+1\right]$ $\frac12\left[\sqrt{10-2\sqrt5}+\sqrt3(-1+\sqrt5)\right]$ 24 2π/15
67.5 3π/8 $\frac{1}{2} \sqrt{2 + \sqrt{2}}$ $\sqrt{2} + 1$ 22.5 π/8
69 23π/60 $\frac{1}{16}\left[2(\sqrt3-1)\sqrt{5-\sqrt5}+\sqrt2(\sqrt3+1)(1+\sqrt5)\right]$ $\frac{1}{4}\left[2-(2-\sqrt3)(3-\sqrt5)\right]\left[2+\sqrt{2(5+\sqrt5)}\right]$ 21 7π/60
72 2π/5 $\frac{\sqrt{2+\phi}}{2}=\frac{\sqrt{5+\sqrt5}}{2\sqrt2}$ $\sqrt{5+2\sqrt5}$ 18 π/10
75 5π/12 $\frac{\sqrt{2+\sqrt{3}}}{2}$ $2+\sqrt{3}$ 15 π/12
78 13π/30 $\frac{1}{8} \left[\sqrt{30+6\sqrt5}+\sqrt5-1\right]$ $\frac12\left[\sqrt{10+2\sqrt5}+\sqrt3(1+\sqrt5)\right]$ 12 π/15
81 9π/20 $\frac12\sqrt{2+\sqrt{2+\phi}}=\frac12\sqrt{2+\sqrt{\frac{5+\sqrt5}{2}}}$ $1+\sqrt5+\sqrt{5+2\sqrt5}$ 9 π/20
84 7π/15 $\frac{1}{8} \left[\sqrt{10-2\sqrt5}+\sqrt3(1+\sqrt5)\right]$ $\frac12\left[\sqrt3(3+\sqrt5)+\sqrt{2(25+11\sqrt5)}\right]$ 6 π/30
87 29π/30 $\frac{1}{16}\left[(2+2\sqrt{3})\sqrt{5+\sqrt{5}}+\sqrt{2}(-1+\sqrt{5})(-1+\sqrt{3})\right]$ $\frac14\left[(2+\sqrt{3})(3+\sqrt{5})-2\right]\left[2+\sqrt{10-2\sqrt{5}}\right]$ 3 π/60
90 π/2 1 $\infty$ 0 0
deg rad sin(a) = cos(b) = sin(90-b) = cos(90-a) tan(a) = cot(b) = tan(90-b) = cot(90-a) deg rad
a b

The values for angles outside the range 0-90 degrees, can be found by reducing the angle be in the correct range. Negative angles should have 360 added to them until positive. If angle ≥360, subtract multiples of 360 from it, until <360. Then, if angle ≥270, subtract from 360. If angle ≥180, subtract 180. If angle ≥90, subtract from 180.

Examples:

• sin(-54)=sin(360-54)=sin(306)=-sin(54)
• cos(120)=cos(180-120)=-cos(60)
• tan(225)=tan(225-180)=tan(45)
• tan(-225)=tan(360-225)=tan(135)=tan(180-135)=-tan(45)
• sin(1000)=sin(1000-360*2)=sin(280)=sin(360-280)=-sin(80)

Angles between 0 and 360 will have the following signs and are measured in a counter-clockwise arc originating at the “x+” line.

FYI, here is the exact value of sin 1°, where $i=\sqrt{-1}$,

$\frac12(-1-i\sqrt3)\sqrt[3]{-\frac{\sqrt6}{384} (\sqrt5-1)(3+\sqrt3)+ \frac{\sqrt3}{192}(3-\sqrt3)\sqrt{5+\sqrt5}+\frac{i}{8}\sqrt{1-\frac1{48}\left(\sqrt6(\sqrt5-1)(3+\sqrt3)-2\sqrt3(3-\sqrt3)\sqrt{5+\sqrt5} \right)^2} } + \\ \frac12(-1+i\sqrt3)\sqrt[3]{-\frac{\sqrt6}{384} (\sqrt5-1)(3+\sqrt3)+ \frac{\sqrt3}{192}(3-\sqrt3)\sqrt{5+\sqrt5}-\frac{i}{8}\sqrt{1-\frac1{48}\left(\sqrt6(\sqrt5-1)(3+\sqrt3)-2\sqrt3(3-\sqrt3)\sqrt{5+\sqrt5} \right)^2} }$

and the cos 1°,

$\frac12\sqrt{2- (i\sqrt3-1)\sqrt[3]{\frac{-\sqrt3}{64}(1+\sqrt5)-\frac{\sqrt2}{128}(\sqrt5-1)\sqrt{5+\sqrt5} + \frac{i}{8}\sqrt{1-\left[\frac1{16}(2\sqrt3(1+\sqrt5)+\sqrt2(\sqrt5-1)\sqrt{5+\sqrt5}\right]^2}} + (i\sqrt3+1)\sqrt[3]{\frac{-\sqrt3}{64}(1+\sqrt5)-\frac{\sqrt2}{128}(\sqrt5-1)\sqrt{5+\sqrt5} – \frac{i}{8}\sqrt{1-\left[\frac1{16}(2\sqrt3(1+\sqrt5)+\sqrt2(\sqrt5-1)\sqrt{5+\sqrt5}\right]^2}} }$

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