# Calculus/Table of Trigonometry

 ← References Calculus Summation notation → Table of Trigonometry

## Definitions

• ${\displaystyle \tan(x)={\frac {\sin(x)}{\cos(x)}}}$
• ${\displaystyle \sec(x)={\frac {1}{\cos(x)}}}$
• ${\displaystyle \cot(x)={\frac {\cos(x)}{\sin(x)}}={\frac {1}{\tan(x)}}}$
• ${\displaystyle \csc(x)={\frac {1}{\sin(x)}}}$

## Inverse trigonometric functions

• ${\displaystyle \arcsin(x)=\int _{0}^{x}{\frac {1}{\sqrt {1-t^{2}}}}\mathrm {d} t=-i\log(ix+{\sqrt {1-x^{2}}})}$
• ${\displaystyle \arccos(x)={\frac {\pi }{2}}-\arcsin(x)={\frac {\pi }{2}}-\int _{0}^{x}{\frac {1}{\sqrt {1-t^{2}}}}\mathrm {d} t={\frac {\pi }{2}}+i\log(ix+{\sqrt {1-x^{2}}})}$
• ${\displaystyle \arctan(x)=\int _{0}^{x}{\frac {1}{1+t^{2}}}\mathrm {d} t={\frac {i}{2}}\log \left({\frac {1-ix}{1+ix}}\right)}$
• ${\displaystyle \operatorname {arccsc}(x)=\arcsin \left({\frac {1}{x}}\right)=-i\log \left({\frac {i}{x}}+{\sqrt {1-{\frac {1}{z^{2}}}}}\right)}$
• ${\displaystyle \operatorname {arcsec}(x)=\arccos \left({\frac {1}{x}}\right)={\frac {\pi }{2}}-\arcsin \left({\frac {1}{x}}\right)={\frac {\pi }{2}}+i\log \left({\frac {i}{x}}+{\sqrt {1-{\frac {1}{z^{2}}}}}\right)}$
• ${\displaystyle \operatorname {arccot}(x)=\arctan \left({\frac {1}{x}}\right)={\frac {\pi }{2}}-\arctan(x)={\frac {\pi }{2}}+{\frac {i}{2}}\log \left({\frac {1+ix}{1-ix}}\right)}$
• ${\displaystyle \arcsin(x)+\arcsin(y)=\arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right)}$
• ${\displaystyle \arccos(x)+\arccos(y)=\arccos \left(xy-{\sqrt {(1-x^{2})(1-y^{2})}}\right)}$
• ${\displaystyle \arctan(x)+\arctan(y)=\arctan \left({\frac {x+y}{1-xy}}\right){\pmod {\pi }}}$

## Pythagorean Identities

• ${\displaystyle \sin ^{2}(x)+\cos ^{2}(x)=1}$
• ${\displaystyle 1+\tan ^{2}(x)=\sec ^{2}(x)}$
• ${\displaystyle 1+\cot ^{2}(x)=\csc ^{2}(x)}$

## Double Angle Identities

• ${\displaystyle \sin(2x)=2\sin(x)\cos(x)}$
• ${\displaystyle \cos(2x)=\cos ^{2}(x)-\sin ^{2}(x)}$
• ${\displaystyle \tan(2x)={\frac {2\tan(x)}{1-\tan ^{2}(x)}}}$
• ${\displaystyle \cos ^{2}(x)={\frac {1+\cos(2x)}{2}}}$
• ${\displaystyle \sin ^{2}(x)={\frac {1-\cos(2x)}{2}}}$

## Angle Sum Identities

${\displaystyle \sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)}$
${\displaystyle \sin(x-y)=\sin(x)\cos(y)-\cos(x)\sin(y)}$
${\displaystyle \cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)}$
${\displaystyle \cos(x-y)=\cos(x)\cos(y)+\sin(x)\sin(y)}$
${\displaystyle \sin(x)+\sin(y)=2\sin \left({\frac {x+y}{2}}\right)\cos \left({\frac {x-y}{2}}\right)}$
${\displaystyle \sin(x)-\sin(y)=2\cos \left({\frac {x+y}{2}}\right)\sin \left({\frac {x-y}{2}}\right)}$
${\displaystyle \cos(x)+\cos(y)=2\cos \left({\frac {x+y}{2}}\right)\cos \left({\frac {x-y}{2}}\right)}$
${\displaystyle \cos(x)-\cos(y)=-2\sin \left({\frac {x+y}{2}}\right)\sin \left({\frac {x-y}{2}}\right)}$
${\displaystyle \tan(x)+\tan(y)={\frac {\sin(x+y)}{\cos(x)\cos(y)}}}$
${\displaystyle \tan(x)-\tan(y)={\frac {\sin(x-y)}{\cos(x)\cos(y)}}}$
${\displaystyle \cot(x)+\cot(y)={\frac {\sin(x+y)}{\sin(x)\sin(y)}}}$
${\displaystyle \cot(x)-\cot(y)={\frac {-\sin(x-y)}{\sin(x)\sin(y)}}}$

## Product-to-sum identities

${\displaystyle \cos(x)\cos(y)={\frac {\cos(x+y)+\cos(x-y)}{2}}}$
${\displaystyle \sin(x)\sin(y)={\frac {\cos(x-y)-\cos(x+y)}{2}}}$
${\displaystyle \sin(x)\cos(y)={\frac {\sin(x+y)+\sin(x-y)}{2}}}$
${\displaystyle \cos(x)\sin(y)={\frac {\sin(x+y)-\sin(x-y)}{2}}}$

## In terms of the complex exponential

${\displaystyle e^{i\theta }=\mathrm {cis} \theta =i\sin \theta +\cos \theta }$
${\displaystyle \sin \theta =\mathrm {Re} (e^{i\theta })={\frac {e^{i\theta }-e^{-i\theta }}{2i}}}$
${\displaystyle \cos \theta =\mathrm {Im} (e^{i\theta })={\frac {e^{i\theta }+e^{-i\theta }}{2}}}$
${\displaystyle \tan \theta ={\frac {\sin \theta }{\cos \theta }}={\frac {e^{2i\theta }-1}{i(e^{2i\theta }+1)}}}$
${\displaystyle \csc \theta ={\frac {1}{\sin \theta }}={\frac {2i}{e^{i\theta }-e^{-i\theta }}}}$
${\displaystyle \sec \theta ={\frac {1}{\cos \theta }}={\frac {2}{e^{i\theta }+e^{-i\theta }}}}$
${\displaystyle \cot \theta ={\frac {1}{\tan \theta }}={\frac {i(e^{2i\theta }+1)}{e^{2i\theta }-1}}}$

## Hyperbolic functions

${\displaystyle e^{x}=\sinh x+\cosh x}$
${\displaystyle \cosh ^{2}x-\sinh ^{2}x=1}$
${\displaystyle \mathrm {sech} ^{2}x=1-\tanh ^{2}x}$
${\displaystyle \mathrm {csch} ^{2}x=\mathrm {coth} ^{2}x-1}$
${\displaystyle \sinh x=-i\sin ix={\frac {e^{x}-e^{-x}}{2}}}$
${\displaystyle \cosh x=\cos ix={\frac {e^{x}+e^{-x}}{2}}}$
${\displaystyle \tanh x=-i\tan ix={\frac {e^{x}-e^{-x}}{e^{x}+e^{-x}}}}$
${\displaystyle \mathrm {csch} x=i\csc ix={\frac {2}{e^{x}-e^{-x}}}}$
${\displaystyle \mathrm {sech} x=\sec ix={\frac {2}{e^{x}+e^{-x}}}}$
${\displaystyle \mathrm {coth} x=i\cot ix={\frac {e^{x}+e^{-x}}{e^{x}-e^{-x}}}}$

### Inverses

${\displaystyle \mathrm {arsinh} x=\int _{0}^{x}{\frac {1}{\sqrt {t^{2}+1}}}\mathrm {d} t=\log \left(x+{\sqrt {x^{2}+1}}\right)}$
${\displaystyle \mathrm {arcosh} x=\int _{1}^{x}{\frac {1}{\sqrt {t^{2}-1}}}\mathrm {d} t=\log \left(x+{\sqrt {x^{2}-1}}\right)}$
${\displaystyle \mathrm {artanh} x=\int _{0}^{x}{\frac {1}{1-t^{2}}}\mathrm {d} t={\frac {1}{2}}\log \left({\frac {1+x}{1-x}}\right)}$
${\displaystyle \mathrm {arccsh} x=\log \left({\frac {1+{\sqrt {1+x^{2}}}}{x}}\right)}$
${\displaystyle \mathrm {arsech} x=\log \left({\frac {1+{\sqrt {1-x^{2}}}}{x}}\right)}$
${\displaystyle \mathrm {arcoth} x={\frac {1}{2}}\log \left({\frac {x+1}{x-1}}\right)}$