微積分/三角函式表

定義

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

反三角函式

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

勾股定理恆等式

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

倍角恆等式

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

角和恆等式

$\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$

$\sin(x-y)=\sin(x)\cos(y)-\cos(x)\sin(y)$

$\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$

$\cos(x-y)=\cos(x)\cos(y)+\sin(x)\sin(y)$

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

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

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

$\tan(x)+\tan(y)={\frac {\sin(x+y)}{\cos(x)\cos(y)}}$

$\tan(x)-\tan(y)={\frac {\sin(x-y)}{\cos(x)\cos(y)}}$

$\cot(x)+\cot(y)={\frac {\sin(x+y)}{\sin(x)\sin(y)}}$

$\cot(x)-\cot(y)={\frac {-\sin(x-y)}{\sin(x)\sin(y)}}$

積化和差公式

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

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

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

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

用復指數表示

$e^{i\theta }=\mathrm {cis} \theta =i\sin \theta +\cos \theta$

$\sin \theta =\mathrm {Re} (e^{i\theta })={\frac {e^{i\theta }-e^{-i\theta }}{2i}}$

$\cos \theta =\mathrm {Im} (e^{i\theta })={\frac {e^{i\theta }+e^{-i\theta }}{2}}$

$\tan \theta ={\frac {\sin \theta }{\cos \theta }}={\frac {e^{2i\theta }-1}{i(e^{2i\theta }+1)}}$

$\csc \theta ={\frac {1}{\sin \theta }}={\frac {2i}{e^{i\theta }-e^{-i\theta }}}$

$\sec \theta ={\frac {1}{\cos \theta }}={\frac {2}{e^{i\theta }+e^{-i\theta }}}$

$\cot \theta ={\frac {1}{\tan \theta }}={\frac {i(e^{2i\theta }+1)}{e^{2i\theta }-1}}$

雙曲函式

$e^{x}=\sinh x+\cosh x$

$\cosh ^{2}x-\sinh ^{2}x=1$

$\mathrm {sech} ^{2}x=1-\tanh ^{2}x$

$\mathrm {csch} ^{2}x=\mathrm {coth} ^{2}x-1$

$\sinh x=-i\sin ix={\frac {e^{x}-e^{-x}}{2}}$

$\cosh x=\cos ix={\frac {e^{x}+e^{-x}}{2}}$

$\tanh x=-i\tan ix={\frac {e^{x}-e^{-x}}{e^{x}+e^{-x}}}$

$\mathrm {csch} x=i\csc ix={\frac {2}{e^{x}-e^{-x}}}$

$\mathrm {sech} x=\sec ix={\frac {2}{e^{x}+e^{-x}}}$

$\mathrm {coth} x=i\cot ix={\frac {e^{x}+e^{-x}}{e^{x}-e^{-x}}}$

反函式

$\mathrm {arsinh} x=\int _{0}^{x}{\frac {1}{\sqrt {t^{2}+1}}}\mathrm {d} t=\log \left(x+{\sqrt {x^{2}+1}}\right)$

$\mathrm {arcosh} x=\int _{1}^{x}{\frac {1}{\sqrt {t^{2}-1}}}\mathrm {d} t=\log \left(x+{\sqrt {x^{2}-1}}\right)$

$\mathrm {artanh} x=\int _{0}^{x}{\frac {1}{1-t^{2}}}\mathrm {d} t={\frac {1}{2}}\log \left({\frac {1+x}{1-x}}\right)$

$\mathrm {arccsh} x=\log \left({\frac {1+{\sqrt {1+x^{2}}}}{x}}\right)$

$\mathrm {arsech} x=\log \left({\frac {1+{\sqrt {1-x^{2}}}}{x}}\right)$

$\mathrm {arcoth} x={\frac {1}{2}}\log \left({\frac {x+1}{x-1}}\right)$

另請參閱

• 三角學