If $\cos x=\frac{-\sqrt{15}}{4}$ and $\frac{\pi}{2}
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 $\therefore \sin \theta=\frac{1}{4}$