Question

Найдите значение выражения √128 cos^2 5pi/8- √32

Answer 1

$\sqrt{128}\cdot cos^2\frac{5\pi }{8}-\sqrt{32}=\sqrt{2^7}\cdot \frac{1+cos\frac{10\pi }{8}}{2}-\sqrt{2^5}=\\\\=8\sqrt2\cdot \frac{1+cos(\pi +\frac{\pi}{4})}{2}-4\sqrt2=4\sqrt2\cdot (1+(-cos\frac{\pi}{4})-1)=\\\\=4\sqrt2\cdot (-\frac{\sqrt2}{2})=-4\\\\\\\star \; \; cos^2\alpha =\frac{1+cos\alpha }{2}\; \; \; ;\; \; \; cos(\pi +\alpha )=-cos\alpha \; \; \star$

Answer 2

$\sqrt{128} cos ^{2}  \frac{5 \pi }{8} - \sqrt{32} = \sqrt{32} ( \sqrt{4} cos ^{2}  \frac{5 \pi }{8} -1)=\sqrt{32} ( 2 cos ^{2}  \frac{5 \pi }{8} -1)=  \\ 
\sqrt{32}cos(2* \frac{5 \pi }{8} )=\sqrt{32}cos\frac{5 \pi }{4} = \sqrt{32} * (-\frac{ \sqrt{2} }{2} )=-4$