在弧度制中：一整个圆周=$2\pi$（等于角度制中的$360^\circ$）
$$ \sin^2\theta+\cos^2\theta=1 $$ $$ 1+\tan^2\theta=\frac{1}{\cos^2\theta} $$ $$ \frac{\sin x}{\cos x}=\tan x $$ $$ \cot x=\frac{1}{\tan x} $$ $$ \csc x=\frac{1}{\sin x} $$ $$ \sec x=\frac{1}{\cos x} $$
$\sin{0}=0 \quad \sin{\frac{\pi}{6}=\frac{1}{2} \quad \sin{\frac{\pi}{4}}=\frac{\sqrt{2}}{2}} \quad \sin{\frac{\pi}{3}}=\frac{\sqrt{3}}{2} \quad \sin{\frac{\pi}{2}}=1 \quad \sin{\pi}=0 \quad \sin{\frac{3}{2}\pi}=-1 \quad \sin{2\pi}=0$
$ \cos{0}=1 \quad \cos{\frac{\pi}{6}=}\frac{\sqrt{3}}{2} \quad \cos{\frac{\pi}{4}}=\frac{\sqrt{2}}{2} \quad \cos{\frac{\pi}{3}}=\frac{1}{2} \quad \cos{\frac{\pi}{2}=0} \quad \cos{\frac{3}{2}\pi}=0 \quad \cos{2\pi}=1 \quad \cos{\pi}=-1$
最小正周期是$2\pi$
$$ \sin(x+2k\pi)=\sin{x},k\in Z \quad \cos(x+2k\pi)=\cos{x},k\in Z$$
$$ \sin(x+(2k+1)\pi)=-\sin{x},k\in Z \quad \cos(x+(2k+1)\pi)=-\cos{x},k\in Z$$
$$ \sin(\pi-x)=-\sin(-x)=\sin{x}$$
$$ \sin(\frac{\pi}{2}-x)=\cos{x}$$
$$ \sin(\alpha \pm \beta)=\sin(\alpha)\cos(\beta) \pm \cos(\alpha)\sin(\beta) $$ $$ \cos(\alpha \pm \beta)=\cos(\alpha)\cos(\beta) \mp \sin(\alpha)\sin(\beta) $$ $$ \tan(\alpha \pm \beta)=\frac{\tan(\alpha) \pm \tan(\beta)}{1 \mp \tan(\alpha)\tan(\beta)} $$
$$ \sin\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{2}} $$ $$ \cos\frac{\theta}{2}=\pm\sqrt{\frac{1+\cos\theta}{2}} $$ $$ \tan\frac{\theta}{2}=\csc\theta-\cot\theta=\pm\sqrt\frac{1-\cos\theta}{1+\cos\theta}=\frac{\sin\theta}{1+\cos\theta}=\frac{1-\cos\theta}{\sin\theta}=\cdots $$
$$ \sin2\theta=2\sin\theta\cos\theta=\frac{2\tan\theta}{1+\tan^2\theta} $$ $$ \cos2\theta=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1=1-2\sin^2\theta $$ $$ \tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}=\frac{1}{1-\tan\theta}-\frac{1}{1+\tan\theta} $$
设$u=\tan{\frac{x}{2}}$，则有 $$ \sin{x}=\frac{2u}{1+u^2} \quad \cos{x}=\frac{1-u^2}{1+u^2} $$
有错误，待检查 $$ a\sin{\alpha}+b\cos{\alpha}=\sqrt{a^2+b^2}\sin{(\alpha+\Phi)},\tan{Phi}=\frac{b}{a}$$