Question

sin^6 альфа+сos^6 альфа+34*sin^2 2альфа

Answer 1

$sin^6 alpha +cos^6 alpha +frac{3}{4}bulletsin^22 alpha =(sin^2 alpha)^3 +(cos^2 alpha)^3 +frac{3}{4}sin^22 alpha=\\=(frac{1-cos2 alpha }{2})^3+(frac{1+cos2 alpha }{2})^3+frac{3}{4}sin^22 alpha =\\=frac{(1-cos2 alpha )^3}{8}+frac{(1+cos2 alpha)^3 }{8}+frac{3}{4}sin^22 alpha=\\=frac{1-cos^32 alpha-3cos2 alpha (1-cos2 alpha ) +1+cos^32 alpha +3cos2 alpha (1+cos2 alpha )}{8}+frac{3}{4}sin^22 alpha =$

$=frac{2-3cos2 alpha +3cos^22 alpha +3cos2 alpha +3cos^22 alpha }{8}+frac{3}{4}sin^22 alpha =\\=frac{2+6cos^22 alpha }{8}+frac{3}{4}sin^22 alpha=frac{1+3cos^22 alpha }{4}+frac{3}{4}sin^22 alpha =frac{1+3cos^22 alpha +3sin^22 alpha }{4}=\\=frac{1+3(cos^22 alpha +sin^22 alpha )}{4}=frac{1+3bullet1}{4}=frac{4}{4}=1.$