Question

Решите пожалуйста, качественно.

Answer 1

$boxed {sin(-alpha )=-sinalpha ; ; ,; ; cos(-alpha )=cosalpha ; ; ,; ; tg(-alpha )=-tgalpha }\\\3); ; frac{pi}{5}=36^circ in 1; chetvert; ; ; to ; ; ; sinfrac{pi}{5}>0; ; ,\\sin(-frac{pi}{5})=-sinfrac{pi}{5}<0; ; ; to ; ; sinfrac{pi}{5}>-sinfrac{pi}{5}\\4); ; 4, radianaapprox 229,2^circin 3; chetvert; ; to ; ; tg4>0\\tg(-4)=-tg4<0; ; to ; ; ; tg(-4)<tg4\\5); ; frac{pi}{12}=15^circ in 1; chetvert; ; ; to ; ; cosfrac{pi}{12}>0$

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

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