# Table of exact trigonometric functions

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 $$\frac{\sqrt{6}-\sqrt{2}}{4} =\frac{\sqrt{2-\sqrt{3}}}{2}$$ $$2-\sqrt{3}$$ 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 $$\frac12\sqrt{3-\phi}=\sqrt{\frac{5-\sqrt5}{8}}$$ $$\sqrt{5-2\sqrt5}$$ 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

\begin{array}{r|l|l|l}
& deg
& \sin
& \cos
& \tan
\\
\hline 2\pi
& 360
& 0
& 1
& 0
\\
\hline \pi
& 180
& 0
& -1
& 0
\\
\hline \frac{2\pi}{3}
& 120
& \frac{1}{2}\sqrt{3}
& -\frac{1}{2}
& -\sqrt{3}
\\
\hline \frac{\pi}{2}
& 90
& 1
& 0
& \pm\infty
\\
\hline \frac{2\pi}{5}
& 72
& \frac{1}{4}\left(\sqrt{10+2\sqrt{5}}\right)
& \frac{1}{4}\left(\sqrt{5}-1\right)
& \sqrt{5+2\sqrt{5}}
\\
\hline \frac{\pi}{3}
&60
& \frac{1}{2}\sqrt{3}
& \frac{1}{2}
& \sqrt{3}
\\
\hline \frac{\pi}{4}
&45
& \frac{1}{2}\sqrt{2}
& \frac{1}{2}\sqrt{2}
& 1
\\
\hline \frac{2\pi}{9}
& 40
& \frac{i}{2}\left(\sqrt[3]{\frac{-1-\sqrt{-3}}{2}}-\sqrt[3]{\frac{-1+\sqrt{-3}}{2}}\right)
& \frac{1}{2}\left(\sqrt[3]{\frac{-1+\sqrt{-3}}{2}}+\sqrt[3]{\frac{-1-\sqrt{-3}}{2}}\right)
&
\\
\hline \frac{\pi}{5}
& 36
& \frac{1}{4}\left(\sqrt{10-2\sqrt{5}}\right)
& \frac{1}{4}\left(\sqrt{5}+1\right)
& \sqrt{5-2\sqrt{5}}
\\
\hline \frac{\pi}{6}
& 30
& \frac{1}{2}
& \frac{1}{2}\sqrt{3}
& \frac{1}{3}\sqrt{3}
\\
\hline \frac{\pi}{7}
&
& \frac{1}{24}\sqrt{3\left(112-\sqrt[3]{14336+\sqrt{-5549064193}}-\sqrt[3]{14336-\sqrt{-5549064193}}\right)}
& \frac{1}{24}\sqrt{3\left(80+\sqrt[3]{14336+\sqrt{-5549064193}}+\sqrt[3]{14336-\sqrt{-5549064193}}\right)}
&
\\
\hline \frac{2\pi}{15}
& 24
& \frac{1}{8}\left(\sqrt{15}+\sqrt{3}-\sqrt{10-2\sqrt{5}}\right)
& \frac{1}{8}\left(1+\sqrt{5}+\sqrt{30-6\sqrt{5}}\right)
& \frac{1}{2}\left(-3\sqrt{3}-\sqrt{15}+\sqrt{50+22\sqrt{5}}\right)
\\
\hline \frac{\pi}{8}
& 22.5
& \frac{1}{2}\left(\sqrt{2-\sqrt{2}}\right)
& \frac{1}{2}\left(\sqrt{2+\sqrt{2}}\right)
& \sqrt{2}-1
\\
\hline \frac{\pi}{9}
& 20
& \frac{i}{4}\left(\sqrt[3]{4-4\sqrt{-3}}-\sqrt[3]{4+4\sqrt{-3}}\right)
& \frac{1}{4}\left(\sqrt[3]{4+4\sqrt{-3}}+\sqrt[3]{4-4\sqrt{-3}}\right)
&
\\
\hline \frac{\pi}{10}
& 18
& \frac{1}{4}\left(\sqrt{5}-1\right)
& \frac{1}{4}\left(\sqrt{10+2\sqrt{5}}\right)
& \frac{1}{5}\left(\sqrt{25-10\sqrt{5}}\right)
\\
\hline \frac{\pi}{12}
& 15
& \frac{1}{4}\left(\sqrt{6}-\sqrt{2}\right)
& \frac{1}{4}\left(\sqrt{6}+\sqrt{2}\right)
& 2-\sqrt{3}
\\
\hline \frac{\pi}{15}
& 12
& \frac{1}{8}\left[\sqrt{10+2\sqrt5}-\sqrt3(-1+\sqrt5)\right]
& \frac{1}{8} \left[\sqrt{30+6\sqrt5}+\sqrt5-1\right]
&
\\
\hline \frac{\pi}{18}
&10
&
&
&
\\
\hline \frac{\pi}{30}
& 6
& \frac{1}{8} \left[\sqrt{30-6\sqrt5}-\sqrt5-1\right]
& \frac{1}{8} \left[\sqrt{10-2\sqrt5}+\sqrt3(1+\sqrt5)\right]
&
\\
\hline \frac{\pi}{40}
& 4.5
& \frac{1}{2} \sqrt{2-\sqrt{2+\sqrt{\frac{5+\sqrt{5}}{2}}}}
& \frac{1}{2} \sqrt{2+\sqrt{2+\sqrt{\frac{5+\sqrt{5}}{2}}}}
&
\\
\hline \frac{\pi}{45}
& 4
&
&
&
\\
\hline \frac{\pi}{60}
&3
& \frac{1}{16}\left[(2-2\sqrt3)\sqrt{5+\sqrt5}+\sqrt2(-1+\sqrt{5})(1+\sqrt3)\right]
& \frac{1}{16}\left[(2+2\sqrt{3})\sqrt{5+\sqrt{5}}+\sqrt{2}(-1+\sqrt{5})(-1+\sqrt{3})\right]
&
\\
\hline \frac{\pi}{90}
&2
&
&
&
\\
\hline \frac{\pi}{180}
&1
& \frac1{2i}\left[\sqrt[3]{\cos(3^\circ) + i\sin(3^\circ)} – \sqrt[3]{cos(3^\circ) – i\sin(3^\circ)} \right]
&
&
\end{array}

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°, $$\frac1{2i}\left[\sqrt[3]{\cos(3) + i\sin(3)} – \sqrt[3]{cos(3) – i\sin(3)} \right]$$, 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°,

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