Question

Преобразуйте левую часть выражения,так чтобы получилась правая

Answer 1

$Cos2\alpha=m\\\\Cos^{2}\alpha=\frac{1+Cos2\alpha }{2}=\frac{1+m}{2}\\\\Sin^{2}\alpha=\frac{1-Cos2\alpha }{2}=\frac{1-m}{2}\\\\Sin^{8}\alpha +Cos^{8}\alpha=(\frac{1-m}{2})^{4}+( \frac{1+m}{2})^{4} =\\\\=\frac{1}{16}[(1-m)^{4}+(1+m)^{4}]=\\\\=\frac{1}{16}(1-4m+6m^{2} -4m^{3}+m^{4}+1+4m+6m^{2}+4m^{3}+m^{4})=\\\\=\frac{1}{16}(2m^{4}+12m^{2}+2)=\frac{1}{8}(m^{4}+6m^{2}+1)=\\\\=\frac{1}{8}(Cos^{4}2\alpha+6Cos^{2}2\alpha+1)=$

$=\frac{1}{8}[(\frac{1+Cos4\alpha }{2})^{2}+6 (\frac{1+Cos4\alpha }{2})+1]=\\\\=\frac{1}{8}(\frac{1+2Cos4\alpha+Cos^{2}4\alpha}{4}+3(1+Cos4\alpha)+1)=\\\\=\frac{1}{32} (1+2Cos4\alpha+Cos^{2}4\alpha+12+12Cos4\alpha+4)=\\\\=\boxed{\frac{1}{32}(Cos^{2}4\alpha+14Cos4\alpha+17)}$