Вычислить sin 2a если sin a = 3/5 и п/2 <a<пcos a = -4/5 и п<a<3п/2 срочно мне нужны...

Решите задачу:

$a)\sin\alpha=\frac35;\ \ \frac\pi2\ \textless \ \alpha\ \textless \ \pi;\\ b)\cos\alpha=-\frac45;\ \ \pi\ \textless \ \alpha\ \textless \ \frac{3\pi}2;\\ \sin2\alpha-?;\\ a)\sin2\alpha=2\cdot\sin\alpha\cdot\cos\alpha;\\ \frac\pi2\ \textless \ \alpha\ \textless \ \pi==\ \textgreater \ \sin\alpha\ \textgreater \ 0\cup\cos\alpha\ \textless \ 0==\ \textgreater \ \\ \sin\alpha=\frac35,\cos\alpha=-\sqrt{1-\sin^2\alpha}=-\sqrt{1-\left(\frac{3}{5}\right)^2}=-\sqrt{1-\frac{9}{25}}=\\ =-\sqrt{\frac{25-9}{25}}=-\sqrt{\frac{16}{25}}=-\frac45;\\ \sin2\alpha=2\cdot\frac35\cdot\left(-\frac45\right)=-\frac{24}{25};\\$
$a)\sin2\alpha=2\cdot\sin\alpha\cdot\cos\alpha;\\ \pi\ \textless \ \alpha\ \textless \ \frac{3\pi}{2}==\ \textgreater \ \sin\alpha\ \textless \ 0\cup\cos\alpha\ \textless \ 0==\ \textgreater \ \\ \cos\alpha=-\frac45,\sin\alpha=-\sqrt{1-\cos^2\alpha}=-\sqrt{1-\left(-\frac{4}{5}\right)^2}=-\sqrt{1-\frac{16}{25}}=\\ =-\sqrt{\frac{25-16}{25}}=-\sqrt{\frac{9}{25}}=-\frac35;\\ \sin2\alpha=2\cdot\left(-\frac35\right)\cdot\left(-\frac45\right)=\frac{24}{25};\\$