Question

Тригонометрия плииз помогите

Answer 1

$\displaystyle \frac{4\cos^22 \alpha -4\cos^2\alpha +3\sin^2\alpha }{4\cos^2( \frac{5 \pi }{2}-\alpha )-\sin^22(\alpha - \pi ) } = \frac{4\cos2\alpha -4\cos^2\alpha +3\sin^2\alpha }{4\sin^2\alpha -\sin^22\alpha }=\\ \\ = \frac{4(2\cos^2\alpha -1)-4\cos^2\alpha +3(1-\cos^2\alpha ) }{4\sin^2\alpha -(2\sin\alpha \cos\alpha )^2}=\\ \\ = \frac{8\cos^2\alpha -4-4\cos^2\alpha +3-3\cos^2\alpha }{4\sin^2\alpha -4\sin^2\alpha \cos^2\alpha } =\\ \\ = \frac{\cos^2\alpha -1}{4\sin^2\alpha (1-\cos^2\alpha )} = \frac{-\sin^2\alpha}{4\sin^2\alpha\cdot \sin^2\alpha}=-\frac{1}{4\sin^2\alpha}$