5.二倍角公式(2)

$(1)\mathrm{原}\mathrm{式}={\displaystyle \frac{2\sin {36}^{\circ }\cos {36}^{\circ }\cos {72}^{\circ }}{2\sin {36}^{\circ }}}$
$={\displaystyle \frac{2\sin {72}^{\circ }\cos {72}^{\circ }}{4\sin {36}^{\circ }}}={\displaystyle \frac{\sin {144}^{\circ }}{4\sin {36}^{\circ }}}$
$={\displaystyle \frac{\sin {36}^{\circ }}{4\sin {36}^{\circ }}}={\displaystyle \frac{1}{4}}$
和初中所学有没有相似
$(2)\mathrm{原}\mathrm{式}=|\sin 2+\cos 2|+|\sin 2-\cos 2|$
$\because {\displaystyle \frac{\pi }{2}}<2<{\displaystyle \frac{3\pi }{4}}$
$\therefore \sqrt{1+\sin 4}+\sqrt{1-\sin 4}=2\sin 2$
认真思考符号问题

$\sin {\displaystyle \frac{\theta }{2}}$

$\sin 2\theta =-\cos (2\theta +{\displaystyle \frac{\pi }{2}})=-\cos 2(\theta +{\displaystyle \frac{\pi }{4}})$
$=-[1-2{\sin }^2(\theta +{\displaystyle \frac{\pi }{4}})]=-{\displaystyle \frac{1}{9}}$
有几个角度？

$-{\displaystyle \frac{5}{9}}$

${\displaystyle \frac{7}{8}}$