Sinальфа =1/2 ,P/2<альфа <P

Решите задачу:

$a)\;\sin\alpha=\frac12\\\sin^2\alpha+\cos^2\alpha=1\Rightarrow \cos\alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-\frac14}=\sqrt{\frac34}=\pm\frac{\sqrt3}2\\\frac\pi2\ \textless \ \alpha\ \textless \ \pi\Rightarrow \cos\alpha\ \textless \ 0\\\cos\alpha=-\frac{\sqrt3}2\\tg\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac12:\frac{\sqrt3}2=\frac12\cdot\frac2{\sqrt3}=\frac1{\sqrt3}\\\\b)\;\cos2\alpha=\cos^2\alpha-\sin^2\alpha=\frac34-\frac14=\frac24=\frac12$