Question

$\sqrt{32}- \sqrt{128} sin^{2} \frac{9\pi}{8}$

Answer 1

$\sqrt{32} - \sqrt{128} sin^2 \frac{9 \pi }{8} = \sqrt{16*2} - \sqrt{64*2} sin^2 \frac{9 \pi }{8} =4 \sqrt{2} -8 \sqrt{2} sin^2 \frac{9 \pi }{8} = \\ =4 \sqrt{2}(1-2sin^2 \frac{9 \pi }{8} )=4 \sqrt{2}cos \frac{9 \pi }{4}=4 \sqrt{2}* \frac{ \sqrt{2} }{2} =4$

Ответ: $4$