Question

30 баллов помогите​

Answer 1

$223.\;\frac{\cos9\alpha-\cos^33\alpha}{\sin9\alpha+\sin^33\alpha}=\frac{4\cos^33\alpha-3\cos3\alpha-\cos^33\alpha}{3\sin3\alpha-4\sin^33\alpha+\sin^33\alpha}=\frac{3\cos^33\alpha-3\cos3\alpha}{3\sin3\alpha-3\sin^33\alpha}=\\\\=\frac{3\cos3\alpha(\cos^23\alpha-1)}{3\sin3\alpha(1-\sin^2\alpha)}=-\frac{3\cos3\alpha(1-\cos^23\alpha)}{3\sin3\alpha(1-\sin^2\alpha)}=-\frac{\cos3\alpha\sin^23\alpha}{\sin3\alpha\cos^23\alpha}=-\frac{\sin3\alpha}{\cos3\alpha}=-tg3\alpha$

$224.\;\sin20^o(4\cos^220^o-1)=\sin20^o(4\cos^220^o-4+3)=\\=\sin20^o(3-4(1-\cos^220^o))=\sin20^o(3-4\sin^220^o)=\\=3\sin20^o-4\sin^320^o=\sin(3\cdot20^o)=\sin60^o=\frac{\sqrt3}2$