Тригонометрия последние два задания

Решите задачу:

$10. \sqrt3cos\alpha-2cos(\alpha-\pi/6)=\sqrt3cos\alpha-2(cos\alpha*cos\frac{\pi}{6}+\\+sin\alpha*sin\frac{\pi}{6})=\sqrt3cos\alpha-2(\frac{\sqrt3}{2}cos\alpha+\frac{1}{2}sin\alpha)=\\=\sqrt3cos\alpha-\sqrt3cos\alpha-sin\alpha=-sin\alpha$

$11. sin\alpha=\frac{8}{17}, cos\alpha=\sqrt{1-sin^2\alpha}=\sqrt{1-\frac{64}{289}}=\sqrt{\frac{225}{289}}=\frac{15}{17}\\ cos\beta=\frac{4}{5}, sin\beta=\sqrt{1-cos^2\alpha}=\sqrt{1-\frac{16}{25}}=\frac{3}{5}\\ sin(\alpha+\beta)=sin\alpha*cos\beta+cos\alpha*sin\beta\\ sin(\alpha+\beta)=\frac{8}{17}*\frac{4}{5}+\frac{15}{17}*\frac{3}{5}=\frac{77}{85}\\ \alpha+\beta=arcsin\frac{77}{85}$