Trigonometry
Basic formulas
\[ \begin{align} \cos(-a)&=\cos a\\ \sin(-a)&=-\sin a\\ \cos(a+b)&= \cos a\cos b-\sin a\sin b \\ \sin(a+b)&=\sin a\cos b+\sin b\cos a\\ \cos^2 a+\sin^2 a&=1\\ \end{align} \]Tangents and arctangents
\[ \begin{align} \tan(\alpha+\beta)={\tan\alpha+\tan\beta\over1-\tan\alpha\tan\beta}\\ \arctan x+\arctan y = \arctan\left({x+y\over1-xy}\right)\\ \end{align} \]Cosines of integer multiples as cosines
\[ \begin{align} \href{monkey.html}{\cos 2\theta = \cos^2\theta-\sin^2\theta} \\ &=2\cos^2\theta - 1\\ \cos 3\theta &= 4\cos^3\theta-3\cos\theta \\ \cos 4\theta &= 8\cos^4\theta-8\cos^2\theta+1 \\ \cos 5\theta &= 16\cos^5\theta-20\cos^3\theta+5\cos\theta \\ \end{align} \]Sines of integer multiples
\[ \begin{align} \sin2\theta&=2\sin\theta\cos\theta\\ \sin3\theta&=-4\sin^3\theta+3\sin\theta \\ \sin4\theta&=-8\sin^3\theta\cos\theta+4\sin\theta\cos\theta \\ \sin5\theta&=16\sin^5\theta-20\sin^3\theta+5\sin\theta\\ \end{align} \]Fractional values
\[ \begin{align} \cos{\theta\over2}&= \sqrt{{1\over2}(1+\cos\theta)} \\ \sin{\theta\over2}&= \sqrt{{1\over2}(1-\cos\theta)} \\ \end{align} \]Particular values
\[ \begin{align} \cos{0}&=1\\ \sin{0}&=0\\ \cos{\pi}&=-1\\ \sin{\pi}&=0\\ \end{align} \]Halves
\[ \begin{align} \cos{\pi\over 2}&=0\\ \sin{\pi\over 2}&=1\\ \end{align} \]Thirds
\[ \begin{align} \cos{\pi\over3}&={1\over2} \\ \sin{\pi\over3}&={\sqrt3\over4} \\ \cos{2\pi\over3}&=-{1\over2} \\ \sin{2\pi\over3}&={\sqrt3\over4} \\ \end{align} \]Quarters
\[ \begin{align} \cos{\pi\over 4}&=1/{\sqrt 2}\\ \sin{\pi\over 4}&=1/{\sqrt 2}\\ \end{align} \]Fifths
\[ \begin{align} \cos{\pi\over5} &=\cos{36^\circ} &={1+\sqrt5\over4} &={1 \over 2}\phi \\ \cos{2\pi\over5} &=\cos{72^\circ} &={\sqrt5-1\over4} &={1 \over 2}(\phi - 1) \\ \cos{3\pi\over5} &=\cos{108^\circ} &={1-\sqrt5\over4} &={1 \over 2}(1-\phi) \\ \end{align} \] where $\phi=(1+\sqrt 5)/2\approx1.618033$, the "Golden ratio".Tenths
\[ \begin{align} \cos{\pi\over10}&=\cos{18^\circ}&=\sqrt{{5+\sqrt5\over8}} &={1\over 2}\sqrt{2+\phi} \\ \cos{3\pi\over10} &=\cos{54^\circ} &=\sqrt{{5\over8}-{\sqrt5\over8}} \\ \sin{\pi\over10} &=\sin{18^\circ} &={\sqrt 5 - 1\over 4} &= {1 \over 2}(\phi - 1)\\ \end{align} \]Derivatives
\[ {d\over d\theta} \]Integrals
\[ \int{1\over\cos\theta}d\theta = \ln(\tan\theta+\sec\theta) + C \] \[ \int{\tan\theta}d\theta = \ln(\sec\theta) + C \]
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