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
deg rad deg rad
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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