💖同角三角基本关系式

${\sin }^2\alpha +{\cos }^2\alpha =1$
$\tan \alpha ={\displaystyle \frac{\sin \alpha }{\cos \alpha }}$

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

$\mathrm{化}\mathrm{简}\sqrt{{\displaystyle \frac{1+\sin x}{1-\sin x}}},x\mathrm{在}\mathrm{第}\mathrm{三}\mathrm{象}\mathrm{限}$

$\mathrm{已}\mathrm{知}\tan \alpha =2,\mathrm{求}\mathrm{①}{\displaystyle \frac{\sin \alpha +2\cos \alpha }{\sin \alpha -\cos \alpha }}\mathrm{②}\sin \alpha \cos \alpha$
同角三角基本关系式2 💖齐次化方法

$\mathrm{已}\mathrm{知}\sin \alpha +\cos \alpha ={\displaystyle \frac{1}{5}},\alpha \in (0,\pi )，\mathrm{求}：$
$(1)\sin \alpha \cos \alpha$
$(2)\sin \alpha -\cos \alpha$
同角三角基本关系式3

$\mathrm{已}\mathrm{知}\alpha \in (0,{\displaystyle \frac{\pi }{4}}),\sin \alpha \cos \alpha ={\displaystyle \frac{12}{25}},\mathrm{则}$

• $\sin \alpha +\cos \alpha ={\displaystyle \frac{7}{5}}$
• $\sin \alpha -\cos \alpha ={\displaystyle \frac{1}{5}}$
• $\sin \alpha ={\displaystyle \frac{3}{5}}$
• $\tan \alpha ={\displaystyle \frac{3}{4}}$
  基本关系式题目4