Solve this

Solution:

Given: $\cos x=-\frac{\sqrt{15}}{4}$

To find: value of sinx

Given that: $\frac{\pi}{2}

So, x lies in IInd quadrant and sin will be positive.

We know that,

$\cos ^{2} \theta+\sin ^{2} \theta=1$

Putting the values, we get

$\left(-\frac{\sqrt{15}}{4}\right)^{2}+\sin ^{2} \theta=1$ [Given]

$\Rightarrow \frac{15}{16}+\sin ^{2} \theta=1$+

$\Rightarrow \sin ^{2} \theta=1-\frac{15}{16}$

$\Rightarrow \sin ^{2} \theta=\frac{16-15}{16}$

$\Rightarrow \sin ^{2} \theta=\frac{1}{16}$

$\Rightarrow \sin \theta=\sqrt{\frac{1}{16}}$

$\Rightarrow \sin \theta=\pm \frac{1}{4}$

Since, $x$ in II $^{\text {nd }}$ quadrant and $\sin \theta$ is positive in II $^{\text {nd }}$ quadrant

$\therefore \sin \theta=\frac{1}{4}$