Дополнительные тригонометрические тождества
\[{\displaystyle \sin \left({\frac {\pi }{4}}+x\right)=\cos \left({\frac {\pi }{4}}-x\right)}\]
\[{\displaystyle \sin \left({\frac {\pi }{4}}-x\right)=\cos \left({\frac {\pi }{4}}+x\right)} \]
\[1\pm \sin x=2\sin ^{2}\left({\frac {\pi }{4}}\pm {\frac x2}\right). \]
\[1+\cos x=2\cos ^{2}\left({\frac x2}\right). \]
\[1-\cos x=2\sin ^{2}\left({\frac x2}\right). \]
\[\sin ^{2}x={\frac {1}{1+\operatorname {ctg}^{2}x}}. \]
\[\cos ^{2}x={\frac {1}{1+\operatorname {tg}^{2}x}}. \]
\[{\displaystyle \sin ^{2}x-\sin ^{2}y=\sin(x-y)\cdot \sin(x+y)} \]
\[{\displaystyle \cos ^{2}x-\cos ^{2}y=-\sin(x-y)\cdot \sin(x+y)} \]
\[{\displaystyle \cos ^{2}x-\sin ^{2}y=\cos(x-y)\cdot \cos(x+y)} \]
\[{\displaystyle \sin 2x+\sin 2y=2\cos(x-y)\cdot \sin(x+y)} \]
\[{\displaystyle \sin 2x-\sin 2y=2\sin(x-y)\cdot \cos(x+y)} \]
\[{\displaystyle \cos 2x+\cos 2y=2\cos(x-y)\cdot \cos(x+y)} \]
\[{\displaystyle \cos 2x-\cos 2y=-2\sin(x-y)\cdot \sin(x+y)} \]
\[{\displaystyle \sin 2x+\cos 2y=2\sin \left({\frac {\pi }{4}}+x-y\right)\cdot \sin \left({\frac {\pi }{4}}+x+y\right)} \]
\[{\displaystyle \sin 2x-\cos 2y=-2\sin \left({\frac {\pi }{4}}-x-y\right)\cdot \sin \left({\frac {\pi }{4}}-x+y\right)} \]
\[{\displaystyle \sin ^{3}x+\cos ^{3}x=(\sin x+\cos x)(1-\sin x\cos x)} \]
\[{\displaystyle \sin ^{4}x+\cos ^{4}x=1-2\sin ^{2}x\,\cos ^{2}x=1-{\frac {1}{2}}\sin ^{2}(2x)={\frac {3}{4}}+{\frac {1}{4}}\cos(4x)} \]
\[{\displaystyle \sin ^{6}x+\cos ^{6}x=1-3\sin ^{2}x\,\cos ^{2}x=1-3\sin ^{2}x+3\sin ^{4}x=1-{\frac {3}{4}}\sin ^{2}(2x)={\frac {5}{8}}+{\frac {3}{8}}\cos(4x)} \]
\[1\pm \operatorname {tg}x={\frac {{\sqrt {2}}\sin \left({\frac {\pi }{4}}\pm x\right)}{\cos x}}. \]
\[1\pm \operatorname {ctg}x={\frac {{\sqrt {2}}\sin \left({\frac {\pi }{4}}\pm x\right)}{\sin x}}. \]
\[\operatorname {tg}x={\frac {\sin 2x}{\cos 2x+1}}={\frac {1-\cos 2x}{\sin 2x}}. \]
\[{\displaystyle \operatorname {ctg} ^{2}x-\operatorname {tg} ^{2}x={\frac {4\cos 2x}{\sin ^{2}2x}}} \]
\[{\displaystyle \sin 3x=4\sin x\cdot \sin \left({\frac {\pi }{3}}+x\right)\cdot \sin \left({\frac {\pi }{3}}-x\right)} \]
\[\operatorname {tg}3x=\operatorname {tg}x\cdot \operatorname {tg}\left({\frac {\pi }{3}}+x\right)\cdot \operatorname {tg}\left({\frac {\pi }{3}}-x\right). \]
\[\sin 5x=16\sin x\cdot \sin \left({\frac {\pi }{5}}+x\right)\cdot \sin \left({\frac {\pi }{5}}-x\right)\cdot \sin \left({\frac {2\pi }{5}}+x\right)\cdot \sin \left({\frac {2\pi }{5}}-x\right) \]
\[\operatorname {tg}5x=\operatorname {tg}x\cdot \operatorname {tg}\left({\frac {\pi }{5}}+x\right)\cdot \operatorname {tg}\left({\frac {\pi }{5}}-x\right)\cdot \operatorname {tg}\left({\frac {2\pi }{5}}+x\right)\cdot \operatorname {tg}\left({\frac {2\pi }{5}}-x\right). \]
\[\sin 7x=64\sin x\cdot \sin \left({\frac {\pi }{7}}+x\right)\cdot \sin \left({\frac {\pi }{7}}-x\right)\cdot \sin \left({\frac {2\pi }{7}}+x\right)\cdot \sin \left({\frac {2\pi }{7}}-x\right)\cdot \sin \left({\frac {3\pi }{7}}+x\right)\cdot \sin \left({\frac {3\pi }{7}}-x\right). \]
\[\operatorname {tg}7x=\operatorname {tg}x\cdot \operatorname {tg}\left({\frac {\pi }{7}}+x\right)\cdot \operatorname {tg}\left({\frac {\pi }{7}}-x\right)\cdot \operatorname {tg}\left({\frac {2\pi }{7}}+x\right)\cdot \operatorname {tg}\left({\frac {2\pi }{7}}-x\right)\cdot \operatorname {tg}\left({\frac {3\pi }{7}}+x\right)\cdot \operatorname {tg}\left({\frac {3\pi }{7}}-x\right). \]
\[\sin nx=2^{{n-1}}\prod \limits _{{k=0}}^{{n-1}}\sin \left(x+{\frac {\pi k}{n}}\right)\]
\[\operatorname {tg}{\big [}(2n+1)x{\big ]}=(-1)^{n}\prod \limits _{{k=0}}^{{2n}}\operatorname {tg}\left(x+{\frac {\pi k}{2n+1}}\right)\]
\[{\displaystyle \sum \limits _{k=1}^{n}\sin(kx)=\sin \left({\frac {n+1}{2}}x\right){\frac {\sin \left({\frac {nx}{2}}\right)}{\sin \left({\frac {x}{2}}\right)}}} \]
\[{\displaystyle \sum \limits _{k=1}^{n}\cos(kx)=\cos \left({\frac {n+1}{2}}x\right){\frac {\sin \left({\frac {nx}{2}}\right)}{\sin \left({\frac {x}{2}}\right)}}} \]
\[\sum \limits _{{k=1}}^{{n}}\sin {\big [}(2k-1)x{\big ]}={\frac {\sin ^{2}nx}{\sin x}}\]
\[\sum \limits _{{k=1}}^{{n}}\cos {\big [}(2k-1)x{\big ]}={\frac {\sin 2nx}{2\sin x}}\]
\[{\displaystyle \sum \limits _{k=1}^{n}\cos {\frac {2\pi k}{2n+1}}=-{\frac {1}{2}}} \]
\[{\displaystyle \prod \limits _{k=1}^{n-1}\sin {\frac {k\pi }{n}}={\frac {n}{2^{n-1}}}} \]
\[{\displaystyle \prod \limits _{k=1}^{n}\sin {\frac {k\pi }{2(n+1)}}={\frac {\sqrt {n+1}}{2^{n}}}} \]
\[{\displaystyle \prod \limits _{k=1}^{n}\sin {\frac {k\pi }{2n+1}}={\frac {\sqrt {2n+1}}{2^{n}}}} \]
\[{\displaystyle \prod \limits _{k=1}^{2n-1}\cos {\frac {k\pi }{n}}={\frac {(-1)^{n}-1}{2^{2n-1}}}} \]
\[\prod \limits _{{k=0}}^{n}\cos \left(2^{k}x\right)={\frac {\sin \left(2^{{n+1}}x\right)}{2^{{n+1}}\sin x}}\]
\[{\displaystyle \prod \limits _{k=0}^{n}\cos {\frac {x}{2^{k}}}={\frac {\sin 2x}{2^{n+1}\sin \left({\frac {x}{2^{n}}}\right)}}} \]
\[{\displaystyle \prod \limits _{k=1}^{n}\cos {\frac {x}{2^{k}}}={\frac {\sin x}{2^{n}\sin \left({\frac {x}{2^{n}}}\right)}}} \]
\[{\displaystyle \prod \limits _{k=0}^{\infty }\cos {\frac {x}{2^{k}}}={\frac {\sin 2x}{2x}}} \]
\[{\displaystyle \prod \limits _{k=1}^{\infty }\cos {\frac {x}{2^{k}}}={\frac {\sin x}{x}}} \]
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