Question

ОБЧИСЛИТИ(У ГРАДУСАХ)
arccos(sin 40π 6) - 3 arctg( tg 40π 6 )

Answer 1

$1); sinfrac{40pi}{6}=sinfrac{20pi}{3}=sin(6pi +frac{2pi}{3})=sinfrac{2pi}{3}=sin (frac{pi}{2}+frac{pi}{6})=cosfrac{pi}{6}\\2); arccos(sinfrac{40pi}{6})=arccos(cosfrac{pi}{6})=arccosfrac{sqrt3}{2}=frac{pi}{6}\\3); tgfrac{40pi}{6}=tg(6pi +frac{2pi}{3})=tgfrac{2pi}{3}=tg(pi -frac{pi}{3})=-tgfrac{pi}{3}\\4); arctg(tgfrac{40pi}{6})=arctg(-tgfrac{pi}{3})=-arctg(tgfrac{pi }{3})=-arctg(sqrt3)=-frac{pi}{3}$

$5); ; arccos(sinfrac{40pi}{6})-3arctg(tgfrac{40pi}{6})=frac{pi}{6}-3cdot (-frac{pi}{3})=\\=frac{pi}{6}+pi =frac{7pi}{6}=7cdot frac{pi}{6}=7cdot 30^{circ }=210^{circ}$