Question

Напишите решение 189,194.

Answer 1

$3( \frac{2}{Sin6 \alpha }-tg3 \alpha )=3( \frac{2}{2Sin3 \alpha Cos3 \alpha }- \frac{Sin3 \alpha }{Cos3 \alpha } )=3( \frac{1}{Sin3 \alpha Cos3 \alpha }- \frac{Sin3 \alpha }{Cos3 \alpha })$$=3* \frac{1-Sin ^{2} 3 \alpha }{Sin3 \alpha Cos3 \alpha } =3* \frac{Cos ^{2}3 \alpha }{Sin3 \alpha Cos3 \alpha }=3* \frac{Cos3 \alpha }{Sin3 \alpha }=3Ctg3 \alpha$

$\frac{Cos ^{5} \alpha -Sin ^{4} \alpha Cos \alpha }{Cos ^{2} \alpha -Sin ^{2} \alpha } -Cos \alpha = \frac{Cos \alpha (Cos ^{4} \alpha -Sin ^{4} \alpha ) }{Cos ^{2} \alpha -Sin ^{2} \alpha }-Cos \alpha =$$= \frac{Cos \alpha (Cos ^{2} \alpha -Sin ^{2} \alpha )(Cos ^{2} \alpha +Sin ^{2} \alpha ) }{Cos ^{2} \alpha -Sin ^{2} \alpha }-Cos \alpha =Cos \alpha *1-Cos \alpha =$$=Cos \alpha -Cos \alpha =0$