3角関数3つでの積和公式・和積公式
3角関数3つでの積和公式・和積公式
3角関数について次が成り立つ。
3角関数について次が成り立つ。
(1)
\[ \sin A+\sin B+\sin C=4\sin\frac{B+C}{2}\sin\frac{C+A}{2}\sin\frac{A+B}{2}+\sin\left(A+B+C\right) \](2)
\[ \sin A\sin B\sin C=\frac{1}{4}\left\{ \sin\left(A+B+C\right)+\sin\left(-A+B+C\right)+\sin\left(A-B+C\right)+\sin\left(A+B-C\right)\right\} \](3)
\[ \cos A+\cos B+\cos C=4\cos\frac{B+C}{2}\cos\frac{C+A}{2}\cos\frac{A+B}{2}-\cos\left(A+B+C\right) \](4)
\[ \cos A\cos B\cos C=\frac{1}{4}\left\{ \cos\left(A+B+C\right)+\cos\left(-A+B+C\right)+\cos\left(A-B+C\right)+\cos\left(A+B-C\right)\right\} \](5)
\[ \tan\left(A+B+C\right)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A} \](6)
\[ \tan^{-1}\left(A+B+C\right)=\frac{\tan^{-1}A+\tan^{-1}B+\tan^{-1}C-\tan^{-1}A\tan^{-1}B\tan^{-1}C}{1-\tan^{-1}A\tan^{-1}B-\tan^{-1}B\tan^{-1}C-\tan^{-1}C\tan^{-1}A} \](1)
\begin{align*} \sin A+\sin B+\sin C & =\sin A+\sin B+\sin C-\sin\left(A+B+C\right)+\sin\left(A+B+C\right)\\ & =2\sin\frac{A+B}{2}\cos\frac{A-B}{2}+\sin\left(-\frac{A+B}{2}\right)\cos\frac{A+B+2C}{2}+\sin\left(A+B+C\right)\\ & =2\sin\frac{A+B}{2}\left(\cos\frac{A-B}{2}-\cos\frac{A+B+2C}{2}\right)+\sin\left(A+B+C\right)\\ & =2\sin\frac{A+B}{2}\left\{ -2\sin\frac{A+C}{2}\sin\left(-\frac{B+C}{2}\right)\right\} +\sin\left(A+B+C\right)\\ & =4\sin\frac{B+C}{2}\sin\frac{C+A}{2}\sin\frac{A+B}{2}+\sin\left(A+B+C\right) \end{align*}(2)
\begin{align*} \sin A\sin B\sin C & =-\frac{1}{2}\left\{ \cos\left(A+B\right)-\cos\left(A-B\right)\right\} \sin C\\ & =-\frac{1}{2}\left\{ \frac{1}{2}\left\{ \sin\left(A+B+C\right)+\sin\left(C-A-B\right)\right\} -\frac{1}{2}\left\{ \sin\left(A-B+C\right)+\sin\left(C-A+B\right)\right\} \right\} \\ & =\frac{1}{4}\left\{ \sin\left(A+B+C\right)+\sin\left(-A+B+C\right)+\sin\left(A-B+C\right)+\sin\left(A+B-C\right)\right\} \end{align*}(3)
\begin{align*} \cos A+\cos B+\cos C & =\cos A+\cos B+\cos C+\cos\left(A+B+C\right)-\cos\left(A+B+C\right)\\ & =2\cos\frac{A+B}{2}\cos\frac{A-B}{2}+2\cos\frac{A+B+2C}{2}\cos\left(-\frac{A+B}{2}\right)-\cos\left(A+B+C\right)\\ & =2\cos\frac{A+B}{2}\left(\cos\frac{A-B}{2}+\cos\frac{A+B+2C}{2}\right)-\cos\left(A+B+C\right)\\ & =2\cos\frac{A+B}{2}\left(2\cos\frac{A+C}{2}\cos\left(-\frac{B+C}{2}\right)\right)-\cos\left(A+B+C\right)\\ & =4\cos\frac{B+C}{2}\cos\frac{C+A}{2}\cos\frac{A+B}{2}-\cos\left(A+B+C\right) \end{align*}(4)
\begin{align*} \cos A\cos B\cos C & =\frac{1}{2}\left\{ \cos\left(A+B\right)+\cos\left(A-B\right)\right\} \cos C\\ & =\frac{1}{2}\left\{ \frac{1}{2}\left\{ \cos\left(A+B+C\right)+\cos\left(A+B-C\right)\right\} +\frac{1}{2}\left\{ \cos\left(A-B+C\right)+\cos\left(A-B-C\right)\right\} \right\} \\ & =\frac{1}{4}\left\{ \cos\left(A+B+C\right)+\cos\left(-A+B+C\right)+\cos\left(A-B+C\right)+\cos\left(A+B-C\right)\right\} \end{align*}(5)
\begin{align*} \tan\left(A+B+C\right) & =\frac{\tan\left(A+B\right)+\tan C}{1-\tan\left(A+B\right)\tan C}\\ & =\frac{\frac{\tan A+\tan B}{1-\tan A\tan B}+\tan C}{1-\frac{\tan A+\tan B}{1-\tan A\tan B}\tan C}\\ & =\frac{\frac{\tan A+\tan B}{1-\tan A\tan B}+\tan C}{1-\frac{\tan A+\tan B}{1-\tan A\tan B}\tan C}\\ & =\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A} \end{align*}(6)
\begin{align*} \tan^{-1}\left(A+B+C\right) & =\left(\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A}\right)^{-1}\\ & =\frac{1-\tan A\tan B-\tan B\tan C-\tan C\tan A}{\tan A+\tan B+\tan C-\tan A\tan B\tan C}\\ & =\frac{\tan^{-1}A\tan^{-1}B\tan^{-1}C-\tan^{-1}C-\tan^{-1}A-\tan^{-1}B}{\tan^{-1}B\tan^{-1}C+\tan^{-1}C\tan^{-1}A+\tan^{-1}A\tan^{-1}B-1}\\ & =\frac{\tan^{-1}A+\tan^{-1}B+\tan^{-1}C-\tan^{-1}A\tan^{-1}B\tan^{-1}C}{1-\tan^{-1}A\tan^{-1}B-\tan^{-1}B\tan^{-1}C-\tan^{-1}C\tan^{-1}A} \end{align*}
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3角関数・双曲線関数の還元公式(負角・余角・補角)
\[
\sin(-x)=-\sin x
\]
三角関数の合成
\[
a\sin\theta+b\cos\theta =\sqrt{a^{2}+b^{2}}\sin(\theta+\alpha)
\]
逆三角関数と逆双曲線関数の関係
\[
\Sin^{\bullet}\left(iz\right)=i\Sinh^{\bullet}z
\]
3角関数・双曲線関数の無限乗積展開
\[
\sin\left(\pi z\right)=\pi z\prod_{k=1}^{\infty}\left(1-\frac{z^{2}}{k^{2}}\right)
\]