sin 2 θ = 2 sin θ cos θ {\displaystyle \sin 2\theta =2\sin \theta \cos \theta }
cos 2 θ = cos 2 θ − sin 2 θ {\displaystyle \cos 2\theta =\cos ^{2}\theta -\sin ^{2}\theta }
tan 2 θ = 2 tan θ 1 − tan 2 θ {\displaystyle \tan 2\theta ={\frac {2\tan \theta }{1-\tan ^{2}\theta }}}
となることを証明する。
加法定理より、
sin ( α + β ) = sin α cos β + cos α sin β {\displaystyle \sin(\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta }
cos ( α + β ) = cos α cos β − sin α sin β {\displaystyle \cos(\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta }
tan ( α + β ) = tan α + tan β 1 − tan α tan β {\displaystyle \tan(\alpha +\beta )={\frac {\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }}}
したがって、
sin 2 θ = sin ( θ + θ ) = sin θ cos θ + cos θ sin θ = 2 sin θ cos θ {\displaystyle {\begin{aligned}\sin 2\theta &=\sin(\theta +\theta )\\&=\sin \theta \cos \theta +\cos \theta \sin \theta \\&=2\sin \theta \cos \theta \end{aligned}}}
cos 2 θ = cos ( θ + θ ) = cos θ cos θ − sin θ sin θ = cos 2 θ − sin 2 θ {\displaystyle {\begin{aligned}\cos 2\theta &=\cos(\theta +\theta )\\&=\cos \theta \cos \theta -\sin \theta \sin \theta \\&=\cos ^{2}\theta -\sin ^{2}\theta \end{aligned}}}
tan 2 θ = tan ( θ + θ ) = tan θ + tan θ 1 − tan θ tan θ = 2 tan θ 1 − tan 2 θ {\displaystyle {\begin{aligned}\tan 2\theta &=\tan(\theta +\theta )\\&={\frac {\tan \theta +\tan \theta }{1-\tan \theta \tan \theta }}\\&={\frac {2\tan \theta }{1-\tan ^{2}\theta }}\end{aligned}}}
