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

Решите задачу:

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

$\frac{1}{2}(cos0-cos\frac{\pi}{4})-sin0+sin\frac{\pi}{4}+\frac{1}{2}(cos\frac{\pi}{4}+cos0)=\frac{1}{2}(1-\frac{\sqrt{2}}{2})+\frac{\sqrt{2}}{2}+\frac{1}{2}(\frac{\sqrt{2}}{2}+1)=\frac{1}{2}-\frac{\sqrt{2}}{4}+\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{4}+\frac{1}{2}=1+\frac{\sqrt{2}}{2}$