Question

Решите пожалуйста, качественно.

Answer 1

$\boxed {sin(-\alpha )=-sin\alpha \; \; ,\; \; cos(-\alpha )=cos\alpha \; \; ,\; \; tg(-\alpha )=-tg\alpha }\\\\\\3)\; \; \frac{\pi}{5}=36^\circ \in 1\; chetvert\; \; \; \to \; \; \; sin\frac{\pi}{5}>0\; \; ,\\\\sin(-\frac{\pi}{5})=-sin\frac{\pi}{5}<0\; \; \; \to \; \; sin\frac{\pi}{5}>-sin\frac{\pi}{5}\\\\4)\; \; 4\, radiana\approx 229,2^\circ\in 3\; chetvert\; \; \to \; \; tg4>0\\\\tg(-4)=-tg4<0\; \; \to \; \; \; tg(-4)<tg4\\\\5)\; \; \frac{\pi}{12}=15^\circ \in 1\; chetvert\; \; \; \to \; \; cos\frac{\pi}{12}>0$

$cos(-\frac{\pi}{12})=cos\frac{\pi}{12}\\\\6)\; \; tg(-\frac{\pi}{4})+cos(-\frac{\pi}{4})+sin(-\frac{\pi}{4})=-tg\frac{\pi}{4}+cos\frac{\pi}{4}-sin\frac{\pi}{4}=\\\\=-1+\frac{\sqrt2}{2}-\frac{\sqrt2}{2}=1\\\\7)\; \; sin(-\frac{\pi}{6})-cos(-\frac{\pi}{3})-tg(-\frac{\pi}{6})=-sin\frac{\pi}{6}-cos\frac{\pi}{3}+tg\frac{\pi}{6}=\\\\=-\frac{1}{2}-\frac{1}{2}+\frac{\sqrt3}{3}=-1+\frac{\sqrt3}{3}$

$8)\; \; sin(-\frac{3\pi}{2})+cos(-11\pi )=-sin\frac{3\pi}{2}+cos\, 11\pi =\\\\=-sin(\pi+\frac{\pi}{2})+cos(\underbrace {12\pi }_{T=2\pi }-\pi )=-(-sin\frac{\pi}{2})+cos(-\pi )=\\\\=+sin\frac{\pi}{2}+cos\pi =1+(-1)=0\\\\9)\; \; tg(-780^\circ )-ctg(-390^\circ )=-tg780^\circ -(-ctg390^\circ )=\\\\=-tg(\underbrace {4\cdot 180^\circ }_{T=180^\circ }+60^\circ )+ctg(\underbrace {2\cdot 180^\circ }_{T=180^\circ }+30^\circ )=-tg60^\circ +ctg30^\circ =\\\\=-\sqrt3+\sqrt3=0$