Question

√18 - √72sin^2(17П/8)

Answer 1

$\sqrt{9 \cdot 2} - \sqrt{9 \cdot 8 }\cdot (\frac{1-\sin (2 \cdot \frac{17\pi}{8})}{2})=3\sqrt{2} -3 \sqrt{2 \cdot 4} \cdot( \frac{1-\sin ( \frac{17\pi}{4})}{2})=\\ \\ = 3 \cdot (\sqrt{2}-2\sqrt{2} \cdot \frac{1-\sin (4\pi + \frac{\pi}{4})}{2})= 3 \cdot (\sqrt{2}-\sqrt{2} \cdot (1-\sin (\frac{\pi}{4}))=\\ \\ = 3 \cdot (\sqrt{2} - \sqrt{2} \cdot (1 - \frac{\sqrt{2}}{2}))= 3 \cdot (\sqrt{2} - \sqrt{2} + \frac{2}{2})= \\ \\ = 3 \cdot 1=3$