Question

Решите пожалуйста
cos(3п/2+a)*sin^3(п-a)-cos(п+a)*sin^3(3п/2-a)/2sina*cos(2п-a)

Answer 1

$(Cos(\frac{3\pi }{2}+\alpha)*Sin^{3}(\pi-\alpha)-Cos(\pi+\alpha)*Sin^{3}(\frac{3\pi }{2}-\alpha) ):2Sin\alpha Cos(2\pi-\alpha)=\frac{Sin\alpha*Sin^{3}\alpha-Cos\alpha*Cos^{3}\alpha}{2Sin\alpha Cos\alpha}=\frac{Sin^{4}\alpha-Cos^{4}\alpha}{Sin2\alpha}=\frac{(Sin^{2}\alpha+Cos^{2}\alpha)(Sin^{2}\alpha-Cos^{2}\alpha)}{Sin2\alpha}=\frac{1*(-Cos2\alpha)}{Sin2\alpha} =-Ctg2\alpha\\\\Otvet:\boxed{-Ctg2\alpha}$