Question

Найдите значение выражения (sin2x-cos2x)^2 при x=-пи/16 заранее спасибо!

Answer 1

$(sin2x-cos2x)^2=(sinfrac{pi}{8}-cosfrac{pi}{8})^2=sinfrac{pi}{8}sinfrac{pi}{8}-2sinfrac{pi}{8}cosfrac{pi}{8}+cosfrac{pi}{8}cosfrac{pi}{8}=frac{1}{2}(cos(frac{pi}{8}-frac{pi}{8})-cos(frac{pi}{8}+frac{pi}{8}))-sin(frac{pi}{8}+frac{pi}{8})+sin(frac{pi}{8}-frac{pi}{8})+frac{1}{2}(cos(frac{pi}{8}+frac{pi}{8})+cos(frac{pi}{8}-frac{pi}{8}))$

$frac{1}{2}(cos0-cosfrac{pi}{4})-sin0+sinfrac{pi}{4}+frac{1}{2}(cosfrac{pi}{4}+cos0)=frac{1}{2}(1-frac{sqrt{2}}{2})+frac{sqrt{2}}{2}+frac{1}{2}(frac{sqrt{2}}{2}+1)=frac{1}{2}-frac{sqrt{2}}{4}+frac{sqrt{2}}{2}+frac{sqrt{2}}{4}+frac{1}{2}=1+frac{sqrt{2}}{2}$