Тригонометрия. найдите значения выражений

Всё решаем по формулам.

[latex] \sqrt{32}- \sqrt{128}Sin^2 \frac{9 \pi }{8}=4 \sqrt{2}-8 \sqrt{2}Sin^2( \pi + \frac{ \pi }{8})=4 \sqrt{2}-8 \sqrt{2}Sin^2 \frac{ \pi }{8}=[/latex] [latex]=4 \sqrt{2}-8 \sqrt{2}* \frac{1}{2}(1-Cos \frac{ \pi }{4})=4 \sqrt{2}-4 \sqrt{2}(1- \frac{ \sqrt{2}}{2})=4 [/latex] ******* [latex] \sqrt{128}*Cos^2 \frac{7 \pi }{8}- \sqrt{32}=8 \sqrt{2}*Cos^2( \pi - \frac{ \pi }{8})-4 \sqrt{2}= [/latex] [latex]=8 \sqrt{2}Cos^2 \frac{ \pi }{8}-4 \sqrt{2}=8 \sqrt{2}* \frac{1}{2}(1+Cos \frac{ \pi }{4})-4 \sqrt{2}=4 \sqrt{2}+4-4 \sqrt{2}=4 [/latex] ******* [latex]6 \sqrt{2}Cos \frac{ \pi }{4}Cos \frac{7 \pi }{3}=6 \sqrt{2}* \frac{ \sqrt{2} }{2}*Cos(2 \pi + \frac{ \pi }{3})=6Cos \frac{ \pi }{3}=6* \frac{1}{2}=3 [/latex]