三角関数の公式の一覧
tan  ( 2 n + 1 ) θ = tan  θ ⋅ ∑ k = 0 n ( ( − 1 ) k ( 2 n + 1 2 k + 1 ) x k ) ∑ k = 0 n ( ( − 1 ) k ( 2 n + 1 2 k ) x k ) = tan  θ ⋅ ∏ k = 1 n ( tan 2  ( k π 2 n + 1 ) − x ) ∏ k = 1 n ( 1 − tan 2  ( k π 2 n + 1 ) x ) {\displaystyle \tan(2n+1)\theta =\tan \theta \cdot {\frac {\sum _{k=0}^{n}{\biggl (}(-1)^{k}{\binom {2n+1}{2k+1}}x^{k}{\biggr )}}{\sum _{k=0}^{n}{\biggl (}(-1)^{k}{\binom {2n+1}{2k}}x^{k}{\biggr )}}}=\tan \theta \cdot {\frac {\prod _{k=1}^{n}{\Bigl (}\tan ^{2}{\Bigl (}{\frac {k\pi }{2n+1}}{\Bigr )}-x{\Bigr )}}{\prod _{k=1}^{n}{\Bigl (}1-\tan ^{2}{\Bigl (}{\frac {k\pi }{2n+1}}{\Bigr )}x{\Bigr )}}}} = tan  θ ⋅ ∏ k = 1 n ( tan 2  ( k π 2 n + 1 ) − tan 2  θ ) ∏ k = 1 n ( 1 − tan 2  ( k π 2 n + 1 ) tan 2  θ ) {\displaystyle =\tan \theta \cdot {\frac {\prod _{k=1}^{n}{{\Bigl (}\tan ^{2}{\Bigl (}{\frac {k\pi }{2n+1}}{\Bigr )}-\tan ^{2}\theta {\Bigr )}}}{\prod _{k=1}^{n}{\Bigl (}1-\tan ^{2}{\Bigl (}{\frac {k\pi }{2n+1}}{\Bigr )}\tan ^{2}\theta {\Bigr )}}}}
tan  ( θ 1 + θ 2 ) = e 1 e 0 − e 2 = x 1 + x 2 1 − x 1 x 2 = tan  θ 1 + tan  θ 2 1 − tan  θ 1 tan  θ 2 , tan  ( θ 1 + θ 2 + θ 3 ) = e 1 − e 3 e 0 − e 2 = ( x 1 + x 2 + x 3 ) − ( x 1 x 2 x 3 ) 1 − ( x 1 x 2 + x 1 x 3 + x 2 x 3 ) , tan  ( θ 1 + θ 2 + θ 3 + θ 4 ) = e 1 − e 3 e 0 − e 2 + e 4 = ( x 1 + x 2 + x 3 + x 4 ) − ( x 1 x 2 x 3 + x 1 x 2 x 4 + x 1 x 3 x 4 + x 2 x 3 x 4 ) 1 − ( x 1 x 2 + x 1 x 3 + x 1 x 4 + x 2 x 3 + x 2 x 4 + x 3 x 4 ) + ( x 1 x 2 x 3 x 4 ) , {\displaystyle {\begin{aligned}\tan(\theta _{1}+\theta _{2})&={\frac {e_{1}}{e_{0}-e_{2}}}={\frac {x_{1}+x_{2}}{1\ -\ x_{1}x_{2}}}={\frac {\tan \theta _{1}+\tan \theta _{2}}{1\ -\ \tan \theta _{1}\tan \theta _{2}}},\\\\\tan(\theta _{1}+\theta _{2}+\theta _{3})&={\frac {e_{1}-e_{3}}{e_{0}-e_{2}}}={\frac {(x_{1}+x_{2}+x_{3})\ -\ (x_{1}x_{2}x_{3})}{1\ -\ (x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3})}},\\\\\tan(\theta _{1}+\theta _{2}+\theta _{3}+\theta _{4})&={\frac {e_{1}-e_{3}}{e_{0}-e_{2}+e_{4}}}\\\\&={\frac {(x_{1}+x_{2}+x_{3}+x_{4})\ -\ (x_{1}x_{2}x_{3}+x_{1}x_{2}x_{4}+x_{1}x_{3}x_{4}+x_{2}x_{3}x_{4})}{1\ -\ (x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4})\ +\ (x_{1}x_{2}x_{3}x_{4})}},\end{aligned}}}
数学的帰納法を用いて証明が可能である。
上の関係式を cos θ と sin θ で割ると、以下の関係式ができる:
1 + tan 2  θ = sec 2  θ {\displaystyle 1+\tan ^{2}\theta =\sec ^{2}\theta \!}
1 + cot 2  θ = csc 2  θ {\displaystyle 1+\cot ^{2}\theta =\csc ^{2}\theta \!}
これらの式から以下の関係を得る:
tan  β 2 tan  γ 2 + tan  γ 2 tan  α 2 + tan  α 2 tan  β 2 = 1 {\displaystyle \tan {\frac {\beta }{2}}\tan {\frac {\gamma }{2}}+\tan {\frac {\gamma }{2}}\tan {\frac {\alpha }{2}}+\tan {\frac {\alpha }{2}}\tan {\frac {\beta }{2}}=1}
sin  α + sin  β + sin  γ = 4 cos  α 2 cos  β 2 cos  γ 2 {\displaystyle \sin \alpha +\sin \beta +\sin \gamma =4\cos {\frac {\alpha }{2}}\cos {\frac {\beta }{2}}\cos {\frac {\gamma }{2}}}
− sin  α + sin  β + sin  γ = 4 cos  α 2 sin  β 2 sin  γ 2 {\displaystyle -\sin \alpha +\sin \beta +\sin \gamma =4\cos {\frac {\alpha }{2}}\sin {\frac {\beta }{2}}\sin {\frac {\gamma }{2}}}
cos  α + cos  β + cos  γ = 4 sin  α 2 sin  β 2 sin  γ 2 + 1 {\displaystyle \cos \alpha +\cos \beta +\cos \gamma =4\sin {\frac {\alpha }{2}}\sin {\frac {\beta }{2}}\sin {\frac {\gamma }{2}}+1}