Тригонометрические функции

Решите задачу:

$\sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-\frac{225}{289}}=\sqrt{\frac{64}{289}}=\pm\frac8{17}\\\frac{3\pi}2\ \textless \ \alpha\ \textless \ 2\pi\Rightarrow\sin\alpha\ \textless \ 0\\\sin\alpha=-\frac8{17}\\\\\cos2\alpha=\cos^2\alpha-\sin^2\alpha=\frac{225}{289}-\frac{64}{289}=\frac{161}{289}\\\\\cos(\alpha-\frac\pi6)=\cos\alpha\cos\frac\pi6+\sin\alpha\sin\frac\pi6=\frac{15}{17}\cdot\frac{\sqrt3}2+\frac8{17}\cdot\frac12=\frac{15\sqrt3+8}{34}$