三角恒等变换

$\sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta$$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$

$\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta$$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$

$\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$ ($\frac{\sin x}{\cos x}$$\frac{\sin x}{\cos x}$)

0 辅助角公式

$a\sin \alpha +b\cos \alpha =\sqrt{a^2+b^2}\sin (\alpha +\phi )$$a\sin\alpha+b\cos\alpha=\sqrt{a^2+b^2}\sin(\alpha+\varphi)$，其中 $\phi$$\varphi$ 的终边经过$(a,b)$$(a,b)$

1 倍角公式

$\sin 2x=2\sin x\cos x$$\sin 2x=2\sin x\cos x$

$\cos 2x={\cos }^{2}x-{\sin }^{2}x=2{\cos }^{2}x-1=1-2{\sin }^{2}x$$\cos 2x=\cos ^2x-\sin ^2x=2\cos ^2x-1=1-2\sin ^2x$

$\tan 2x=\frac{2\tan x}{1-{\tan }^{2}x}$$\tan 2x=\frac{2\tan x}{1-\tan ^2x}$

2 升幂公式

$1\pm \sin 2x=(\sin x\pm \cos x{)}^2$$1\pm\sin2x=(\sin x\pm\cos x)^2$ _(1-1)

$1+\cos 2x=2{\cos }^{2}x$$1+\cos 2x=2\cos^2x$ _(1-2[3])

$1-\cos 2x=2{\sin }^{2}x$$1-\cos2x=2\sin^2x$ _(1-2[2])

3 降幂公式

${\sin }^2\alpha =\frac{1-\cos 2\alpha }{2}$$\sin^2\alpha=\frac{1-\cos2\alpha}{2}$ _(1-1)

${\cos }^2\alpha =\frac{1+\cos 2\alpha }{2}$$\cos^2\alpha=\frac{1+\cos2\alpha}{2}$ _(2-2)

${\tan }^2\alpha =\frac{1-\cos 2\alpha }{1+\cos 2\alpha }$$\tan^2\alpha=\frac{1-\cos2\alpha}{1+\cos2\alpha}$ _(2-3)

4 万能公式

$\sin 2x=\frac{2\tan x}{1+{\tan }^{2}x}$$\sin2x=\frac{2\tan x}{1+\tan^2x}$ ($\frac{1-1}{{\sin }^2\alpha +{\cos }^2\alpha }$$\frac{1-1}{\sin^2\alpha+\cos^2\alpha}$)

$\cos 2x=\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}$$\cos2x=\frac{1-\tan^2x}{1+\tan^2x}$ ($\frac{1-2[1]}{{\sin }^2\alpha +{\cos }^2\alpha }$$\frac{1-2[1]}{\sin^2\alpha+\cos^2\alpha}$)

$\tan 2x=\frac{2\tan x}{1-{\tan }^{2}x}$$\tan2x=\frac{2\tan x}{1-\tan^2x}$ _(1-3)

5 半角公式

$\sin \frac{x}{2}=\pm \sqrt{\frac{1-\cos x}{2}}$$\sin\frac{x}{2}=\pm\sqrt{\frac{1-\cos x}{2}}$ _(3-3)

$\cos \frac{x}{2}=\pm \sqrt{\frac{1+\cos x}{2}}$$\cos\frac{x}{2}=\pm\sqrt{\frac{1+\cos x}{2}}$ _(3-2)

$\tan \frac{x}{2}=\pm \sqrt{\frac{1-\cos x}{1+\cos x}}$$\tan\frac{x}{2}=\pm\sqrt{\frac{1-\cos x}{1+\cos x}}$ ($\frac{4-1}{4-2}$$\frac{4-1}{4-2}$)

6 积化和差公式

$\cos \alpha \cos \beta ={\displaystyle \frac{1}{2}}[\cos (\alpha +\beta )+\cos (\alpha -\beta )]$$\cos\alpha \cos\beta = \dfrac{1}{2}\big[\cos(\alpha+\beta) + \cos(\alpha-\beta)\big]$ ( $\cos (\alpha +\beta )+\cos (\alpha -\beta )=2\cos \alpha \cos \beta$$\cos(\alpha+\beta)+\cos(\alpha-\beta)=2\cos\alpha\cos\beta$ )

$\sin \alpha \sin \beta ={\displaystyle \frac{1}{2}}[\cos (\alpha -\beta )-\cos (\alpha +\beta )]$$\sin\alpha \sin\beta = \dfrac{1}{2}\big[\cos(\alpha-\beta) - \cos(\alpha+\beta)\big]$ ( $\cos (\alpha -\beta )-\cos (\alpha +\beta )=2\sin \alpha \sin \beta$$\cos(\alpha-\beta)-\cos(\alpha+\beta)=2\sin\alpha\sin\beta$ )

$\sin \alpha \cos \beta ={\displaystyle \frac{1}{2}}[\sin (\alpha +\beta )+\sin (\alpha -\beta )]$$\sin\alpha \cos\beta = \dfrac{1}{2}\big[\sin(\alpha+\beta) + \sin(\alpha-\beta)\big]$ ( $\sin (\alpha +\beta )+\sin (\alpha -\beta )=2\sin \alpha \cos \beta$$\sin(\alpha+\beta)+\sin(\alpha-\beta)=2\sin\alpha\cos\beta$ )

$\cos \alpha \sin \beta ={\displaystyle \frac{1}{2}}[\sin (\alpha +\beta )-\sin (\alpha -\beta )]$$\cos\alpha \sin\beta = \dfrac{1}{2}\big[\sin(\alpha+\beta) - \sin(\alpha-\beta)\big]$ ( $\sin (\alpha +\beta )-\sin (\alpha -\beta )=2\cos \alpha \sin \beta$$\sin(\alpha+\beta)-\sin(\alpha-\beta)=2\cos\alpha\sin\beta$ )

7 和差化积公式

$\cos \alpha +\cos \beta =2\cos {\displaystyle \frac{\alpha +\beta }{2}}\cos {\displaystyle \frac{\alpha -\beta }{2}}$$\cos\alpha + \cos\beta = 2\cos\dfrac{\alpha+\beta}{2} \cos\dfrac{\alpha-\beta}{2}$ ( $2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}=\cos (\frac{\alpha +\beta }{2}+\frac{\alpha -\beta }{2})+\cos (\frac{\alpha +\beta }{2}-\frac{\alpha -\beta }{2})=\cos \alpha +\cos \beta$$2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}=\cos(\frac{\alpha+\beta}{2}+\frac{\alpha-\beta}{2})+\cos(\frac{\alpha+\beta}{2}-\frac{\alpha-\beta}{2})=\cos\alpha+\cos\beta$ )

$\cos \alpha -\cos \beta =-2\sin {\displaystyle \frac{\alpha +\beta }{2}}\sin {\displaystyle \frac{\alpha -\beta }{2}}$$\cos\alpha - \cos\beta = -2\sin\dfrac{\alpha+\beta}{2} \sin\dfrac{\alpha-\beta}{2}$ ( $-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}=\cos (\frac{\alpha +\beta }{2}+\frac{\alpha -\beta }{2})-\cos (\frac{\alpha +\beta }{2}-\frac{\alpha -\beta }{2})=\cos \alpha -\cos \beta$$-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}=\cos(\frac{\alpha+\beta}{2}+\frac{\alpha-\beta}{2})-\cos(\frac{\alpha+\beta}{2}-\frac{\alpha-\beta}{2})=\cos\alpha-\cos\beta$ )

$\sin \alpha +\sin \beta =2\sin {\displaystyle \frac{\alpha +\beta }{2}}\cos {\displaystyle \frac{\alpha -\beta }{2}}$$\sin\alpha + \sin\beta = 2\sin\dfrac{\alpha+\beta}{2} \cos\dfrac{\alpha-\beta}{2}$ ( $2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}=\sin (\frac{\alpha +\beta }{2}+\frac{\alpha -\beta }{2})+\sin (\frac{\alpha +\beta }{2}-\frac{\alpha -\beta }{2})=\sin \alpha +\sin \beta$$2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}=\sin(\frac{\alpha+\beta}{2}+\frac{\alpha-\beta}{2})+\sin(\frac{\alpha+\beta}{2}-\frac{\alpha-\beta}{2})=\sin\alpha+\sin\beta$ )

$\sin \alpha -\sin \beta =2\cos {\displaystyle \frac{\alpha +\beta }{2}}\sin {\displaystyle \frac{\alpha -\beta }{2}}$$\sin\alpha - \sin\beta = 2\cos\dfrac{\alpha+\beta}{2} \sin\dfrac{\alpha-\beta}{2}$ ( $2\cos \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}=\sin (\frac{\alpha +\beta }{2}+\frac{\alpha -\beta }{2})-\sin (\frac{\alpha +\beta }{2}-\frac{\alpha -\beta }{2})=\sin \alpha -\sin \beta$$2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}=\sin(\frac{\alpha+\beta}{2}+\frac{\alpha-\beta}{2})-\sin(\frac{\alpha+\beta}{2}-\frac{\alpha-\beta}{2})=\sin\alpha-\sin\beta$ )