Question

How do you solve this?

Answer 1

$\dfrac{\pi}{2}\leq A\leq\pi\\\\\sin A=\dfrac{3}{5}\\\\\sin2A=2\sin A\cos A\\\\\text{we know}\ \sin^2 x+\cos^2x=1\\\\\text{then}\\\\\left(\dfrac{3}{5}\right)^2+\cos^2A=1\\\\\dfrac{9}{25}+\cos^2A=1\ \ \ \ |-\dfrac{9}{25}\\\\\cos^2A=\dfrac{16}{25}\to\cos A=\pm\sqrt{\dfrac{16}{25}}\\\\\cos A=\pm\dfrac{4}{5}\\(\dfrac{\pi}{2}\leq A\leq\pi\to\cos A < 0)\\\text{therefore}\ \cos A=-\dfrac{4}{5}$

$\text{substitute}\\\\\sin2A=2\sin A\cos A=2\cdot\dfrac{3}{5}\cdot\left(-\dfrac{4}{5}\right)=-\dfrac{24}{25}$