Question

25б!!! Решите:
((sina+sin3a)/(cosa+cos3a))*(1 +cos4a)

Answer 1

$\frac{Sin\alpha+Sin3\alpha}{Cos\alpha+Cos3\alpha}*(1+Cos4\alpha )=\frac{2Sin\frac{\alpha+3\alpha}{2}*Cos\frac{\alpha-3\alpha}{2}}{2Cos\frac{\alpha+3\alpha}{2}*Cos\frac{\alpha-3\alpha}{2}} *(1+Cos4\alpha)=\\\\=\frac{Sin2\alpha*Cos\alpha}{Cos2\alpha*Cos\alpha} *2Cos^{2}2\alpha=\frac{Sin2\alpha*2Cos^{2}2\alpha}{Cos2\alpha }=2Sin2\alpha Cos2\alpha=\boxed{Sin4\alpha}$