Question

помогите умоляю решить пожалуйста!

Answer 1

$Formyla:; ; sin^4 alpha =(sin^2 alpha )^2=(frac{1-cos2 alpha }{2})^2=frac{(1-cos2 alpha )^2}{4}\\sin^4frac{pi}{16}+sin^4frac{3pi}{16}+sin^4frac{5pi}{16}+sin^4frac{7pi}{16}=frac{(1-cosfrac{pi}{8})^2}{4}+frac{(1-cosfrac{3pi}{8})^2}{4}+\\+frac{(1-cosfrac{5pi}{8})^2}{4}+frac{(1-cosfrac{7pi}{8})^2}{4}=frac{1}{4}cdot (1-2cosfrac{pi}{8}+cos^2frac{pi}{8}+\\+1-2cosfrac{3pi}{8}+cos^2frac{3pi}{8}+1-2cosfrac{5pi}{8}+cos^2frac{5pi}{8}+\\+1-2cosfrac{7pi}{8}+$

$+cos^2frac{7pi}{8})=$

$=frac{1}{4}cdot (4-2(cosfrac{pi}{8}+cosfrac{7pi}{8})-2(cosfrac{3pi}{8}+cosfrac{5pi}{8})+cos^2frac{pi}{8}+\\+cos^2frac{3pi}{8}+cos^2frac{5pi}{8}+cos^2frac{7pi}{8})=\\=frac{1}{4}cdot (4-2cdot 2cosfrac{pi}{2}cdot cosfrac{6pi}{8}-2cdot 2cosfrac{pi}{2}cdot cosfrac{2pi}{8}+frac{1+cosfrac{pi}{4}}{2}+\\+frac{1+cosfrac{3pi}{4}}{2}+frac{1+cosfrac{5pi}{4}}{2}+frac{1+cosfrac{7pi}{4}}{2})=; [, cosfrac{pi}{2}=0, ]=$

$=frac{1}{4}cdot (4-0-0+frac{1}{2}cdot (1+frac{sqrt2}{2}+1-frac{sqrt2}{2}+1-frac{sqrt2}{2}+1+frac{sqrt2}{2})=\\=frac{1}{4}cdot (4+frac{1}{2}cdot 4)=frac{6}{4}=frac{3}{2}$