Question

ПОМОГИТЕЕЕЕЕЕЕЕЕЕЕЕЕЕЕЕЕЕЕ!
$8cos(5P/2- \beta )+11sin(7P+ \beta ),sin \beta =1/6$

Answer 1

$8\cos( \frac{5\pi}{2}-\beta)+11\sin(7\pi+\beta)=|\sin\beta= \frac{1}{6}|=\\ =8\cos( \frac{5\pi}{2})\cos\beta+8\sin( \frac{5\pi}{2})\sin\beta+11sin(7\pi)\cos\beta+11\cos(7\pi)sin\beta=\\ =8\cos( \frac{\pi}{2})\cos\beta+8\sin( \frac{\pi}{2})\sin\beta+11sin\pi\cos\beta+11\cos\pi\sin\beta=\\ =8\cdot0\cdot\cos\beta+8\cdot1\cdot\sin\beta+11\cdot0\cdot\cos\beta+11\cdot(-1)\cdot\sin\beta=\\ =8\sin\beta-11\sin\beta=-3\sin\beta=|\sin\beta= \frac{1}{6}|=-3\cdot \frac{1}{6}=- \frac{1}{2}$