三角函数 / trigonometry

基本知识点

primary: $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}},\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}},\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$

secondary: $\csc\theta=\frac{1}{\sin\theta}=\frac{\text{hypotenuse}}{\text{opposite}},\sec\theta=\frac{1}{\cos\theta}=\frac{\text{hypotenuse}}{\text{adjacent}},\cot=\frac{1}{\tan\theta}=\frac{\text{adjacent}}{\text{opposite}}$

基本变换

$\tan\theta=\frac{\sin\theta}{\cos\theta},\cot\theta=\frac{\cos\theta}{\sin\theta}$ $\csc\theta=\frac{1}{\sin\theta},\sec\theta=\frac{1}{\cos\theta},\cot=\frac{1}{\tan\theta}$

$1+\tan^2\theta=\sec^2\theta$, $1+\cot^2\theta=\csc^2\theta$

和差公式

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$
$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$
$\tan(\alpha\pm\beta)=\frac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\beta}$

二倍角公式

$\sin2\theta=2\sin\theta\cos\theta$
$\cos2\theta=2\cos^2\theta-1=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta$
$\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}$

降幂公式

$\sin^2\theta=\frac{1-\cos2\theta}{2}$
$\cos^2\theta=\frac{1+\cos2\theta}{2}$
$\tan^2\theta=\frac{1-\cos2\theta}{1+\cos2\theta}$
$\sin\theta=\pm\sqrt{\frac{1-\cos2\theta}{2}}$
$\cos\theta=\pm\sqrt{\frac{1+\cos2\theta}{2}}$
$\tan\theta=\frac{1-\cos2\theta}{\sin2\theta}=\frac{\sin2\theta}{1+\cos2\theta}$

三倍角公式

$\sin3\theta=\sin\theta\cos2\theta+\cos\theta\sin2\theta=\sin\theta(\cos^2\theta-\sin^2\theta)+2\sin\theta\cos^2\theta=3\sin\theta\cos^2\theta-\sin^3\theta\\=3\sin\theta(1-\sin^2\theta)-\sin^3\theta=\sin\theta(3-4\sin^2\theta)=3\sin\theta-4\sin^3\theta$
$\sin\theta\sin(\frac{\pi}{3}-\theta)\sin(\frac{\pi}{3}+\theta)=\sin\theta(\frac{\sqrt{3}}{2}\cos\theta-\frac{1}{2}\sin\theta)(\frac{\sqrt{3}}{2}\cos\theta+\frac{1}{2}\sin\theta)=\sin\theta(\frac{3}{4}\cos^2\theta-\frac{1}{4}\sin^2\theta)\\=\frac{1}{4}(3\sin\theta\cos^2\theta-\sin^3\theta)=\frac{1}{4}\sin3\theta\Rightarrow\boxed{\sin3\theta=4\sin\theta\sin(\frac{\pi}{3}-\theta)\sin(\frac{\pi}{3}+\theta)}$

$\cos3\theta=\cos\theta\cos2\theta-\sin\theta\sin2\theta=\cos\theta(\cos^2\theta-\sin^2\theta)-2\sin^2\theta\cos\theta=\cos^3\theta-3\sin^2\theta\cos\theta\\=\cos^3\theta-3(1-\cos^2\theta)\cos\theta=\cos\theta(4\cos^2\theta-3)=4\cos^3\theta-3\cos\theta$
$\cos\theta\cos(\frac{\pi}{3}-\theta)\cos(\frac{\pi}{3}+\theta)=\cos\theta(\frac{1}{2}\cos\theta+\frac{\sqrt{3}}{2}\sin\theta)(\frac{1}{2}\cos\theta-\frac{\sqrt{3}}{2}\sin\theta)=\cos\theta(\frac{1}{4}\cos^2\theta-\frac{3}{4}\sin^2\theta)\\=\frac{1}{4}(\cos^3\theta-3\sin^2\theta\cos\theta)=\frac{1}{4}\cos3\theta\Rightarrow\boxed{\cos3\theta=4\cos\theta\cos(\frac{\pi}{3}-\theta)\cos(\frac{\pi}{3}+\theta)}$

$\tan3\theta=\frac{\sin3\theta}{\cos3\theta}=\frac{4\sin\theta\sin(\frac{\pi}{3}-\theta)\sin(\frac{\pi}{3}+\theta)}{4\cos\theta\cos(\frac{\pi}{3}-\theta)\cos(\frac{\pi}{3}+\theta)}=\tan\theta\tan(\frac{\pi}{3}-\theta)\tan(\frac{\pi}{3}+\theta)$

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