trigonometric functions

sin(θ)= $\frac{对边}{斜边}$
cos(θ)= $\frac{邻边}{斜边}$
sin(θ)= $\frac{对边}{邻边}$

证明BF=DF2

$$\{\begin{array}{rl}\angle FOB& =\angle F_{2}OD\\ BO& =DO\\ F_{1}O& =F_{2}O\end{array}\therefore △F_{1}OB=△F_{2}OD$$ $$\therefore F_{1}B=F_{2}D$$

二倍角公式

知道$\tan \alpha$ 求二倍角

$$\sin 2\alpha =2\sin \alpha \cos \alpha =\frac{2\sin \alpha \cos \alpha }{{\sin }^{2}a+{\cos }^2\alpha }(上下同除{\cos }^2\alpha )=\frac{2\tan \alpha }{{\tan }^2\alpha +1}$$ $$\cos 2\alpha ={\cos }^2\alpha -{\sin }^2\alpha =\frac{{\cos }^2\alpha -{\sin }^2\alpha }{{\cos }^2\alpha +{\sin }^2\alpha }=\frac{1-{\tan }^2\alpha }{1+{\tan }^2\alpha }$$

还有一个默认会的。
$\tan 2\alpha =\frac{2\tan \alpha }{1-{\tan }^2\alpha }$

余弦二倍角公式

$\cos 2\alpha =\{\begin{array}{l}{\cos }^2\alpha -{\sin }^2\alpha \\ 2{\cos }^2\alpha -1\\ 1-2{\sin }^2\alpha \end{array}$

$$\begin{array}{r}{\cos }^2\alpha =\frac{1+\cos 2\alpha }{2}\\ {\sin }^2\alpha =\frac{1-\cos 2\alpha }{2}\end{array}$$

诱导公式

$\cos (\frac{7}{2}\pi -\alpha )=\cos (\frac{8}{2}\pi -\frac{1}{2}\pi -\alpha )=\cos (-\frac{\pi }{2}-\alpha )=\cos (\frac{\pi }{2}+\alpha )=\cos (\alpha +\frac{\pi }{2})=-\sin \alpha$

$\cos (-\frac{\pi }{2}-\alpha )=\cos (\frac{\pi }{2}+\alpha )$
利用$f(x)=\cos x$为偶函数的性质。

$\cos (\alpha +\frac{\pi }{2})=-\sin \alpha$
奇变偶不变，符号看象限。$\frac{1}{2}$为奇数变函数名，想象$\alpha$是第一象限的锐角，加上90°(为第二象限)，看$\cos (\alpha +\frac{\pi }{2})$的函数名，在此象限内是正还是负，结果为负。