Question

How do you solve this?

Answer 1

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

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