If $\cos \mathrm{x}=\frac{-3}{5}$ and $\pi<\mathrm{x}<\frac{3 \pi}{2}$ find the values of sin 2x
Given: $\cos x=\frac{-3}{5}$
To find: $\sin 2 x$
We know that,
sin2x = 2 sinx cosx …(i)
Here, we don’t have the value of sin x. So, firstly we have to find the value of sinx
We know that
$\cos ^{2} x+\sin ^{2} x=1$
Putting the values, we get
$\left(-\frac{3}{5}\right)^{2}+\sin ^{2} x=1$
$\Rightarrow \frac{9}{25}+\sin ^{2} x=1$
$\Rightarrow \sin ^{2} x=1-\frac{9}{25}$
$\Rightarrow \sin ^{2} x=\frac{25-9}{25}$
$\Rightarrow \sin ^{2} x=\frac{16}{25}$
$\Rightarrow \sin x=\sqrt{\frac{16}{25}}$
$\Rightarrow \sin x=\pm \frac{4}{5}$
It is given that $\pi $\Rightarrow \sin x=-\frac{4}{5}$ Putting the value of sinx and cosx in eq. (i), we get sin2x = 2sinx cosx $\sin 2 x=2 \times\left(-\frac{4}{5}\right) \times\left(-\frac{3}{5}\right)$ $\therefore \sin 2 \mathrm{x}=\frac{24}{25}$