三角関数の公式の一覧
= tan  θ ⋅ ∏ k = 1 n ( tan  ( k π 2 n + 1 ) + tan  θ ) ⋅ ∏ k = 1 n ( tan  ( k π 2 n + 1 ) − tan  θ ) ∏ k = 1 n ( 1 − tan  ( k π 2 n + 1 ) tan  θ ) ⋅ ∏ k = 1 n ( 1 + tan  ( k π 2 n + 1 ) tan  θ ) {\displaystyle =\tan \theta \cdot {\frac {\prod _{k=1}^{n}{{\Bigl (}\tan {\Bigl (}{\frac {k\pi }{2n+1}}{\Bigr )}+\tan \theta {\Bigr )}}\cdot \prod _{k=1}^{n}{{\Bigl (}\tan {\Bigl (}{\frac {k\pi }{2n+1}}{\Bigr )}-\tan \theta {\Bigr )}}}{\prod _{k=1}^{n}{\Bigl (}1-\tan {\Bigl (}{\frac {k\pi }{2n+1}}{\Bigr )}\tan \theta {\Bigr )}\cdot \prod _{k=1}^{n}{\Bigl (}1+\tan {\Bigl (}{\frac {k\pi }{2n+1}}{\Bigr )}\tan \theta {\Bigr )}}}} = tan  θ ⋅ ∏ k = 1 n ( tan  ( k π 2 n + 1 + θ ) ) ⋅ ∏ k = 1 n ( tan  ( k π 2 n + 1 − θ ) ) {\displaystyle =\tan \theta \cdot {\prod _{k=1}^{n}{{\Biggl (}\tan {\Bigl (}{\frac {k\pi }{2n+1}}+\theta {\Bigr )}{\Biggr )}}\cdot \prod _{k=1}^{n}{{\Biggl (}\tan {\Bigl (}{\frac {k\pi }{2n+1}}-\theta {\Bigr )}{\Biggr )}}}}
( cos  φ − sin  φ sin  φ cos  φ ) ( cos  θ − sin  θ sin  θ cos  θ ) = ( cos  φ cos  θ − sin  φ sin  θ − cos  φ sin  θ − sin  φ cos  θ sin  φ cos  θ + cos  φ sin  θ − sin  φ sin  θ + cos  φ cos  θ ) = ( cos  ( θ + φ ) − sin  ( θ + φ ) sin  ( θ + φ ) cos  ( θ + φ ) ) {\displaystyle {\begin{aligned}&{}\quad \left({\begin{array}{rr}\cos \phi &-\sin \phi \\\sin \phi &\cos \phi \end{array}}\right)\left({\begin{array}{rr}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \end{array}}\right)\\[12pt]&=\left({\begin{array}{rr}\cos \phi \cos \theta -\sin \phi \sin \theta &-\cos \phi \sin \theta -\sin \phi \cos \theta \\\sin \phi \cos \theta +\cos \phi \sin \theta &-\sin \phi \sin \theta +\cos \phi \cos \theta \end{array}}\right)\\[12pt]&=\left({\begin{array}{rr}\cos(\theta +\phi )&-\sin(\theta +\phi )\\\sin(\theta +\phi )&\cos(\theta +\phi )\end{array}}\right)\end{aligned}}}
arcsin  x + arccos  x = π 2 {\displaystyle \arcsin x+\arccos x={\frac {\pi }{2}}}
arctan  x + arccot  x = π 2 . {\displaystyle \arctan x+\operatorname {arccot} x={\frac {\pi }{2}}.}
arctan  x + arctan  1 x = { π 2 , if x > 0 − π 2 , if x < 0 {\displaystyle \arctan x+\arctan {\frac {1}{x}}=\left\{{\begin{matrix}{\frac {\pi }{2}},&{\mbox{if }}x>0\\-{\frac {\pi }{2}},&{\mbox{if }}x<0\end{matrix}}\right.}