Question

The value of $\sin \left[\arccos \left(-\frac{1}{2}\right)\right]$ is -

• A. $\frac{1}{\sqrt{2}}$
• B. 1
• C. $\frac{\sqrt{3}}{1}$
• D. none of these

Answer 1

The correct option is D none of these

$\sin \left(arc\cos (-\frac{1}{2})\right)=\sin \left[{\cos }^{-1}(-\frac{1}{2})\right]$

Let ${\cos }^{-1}(-\frac{1}{2})=P$

$\cos p=-\frac{1}{2}$

$\Rightarrow P=\pi -\frac{\pi }{3}=\frac{2\pi }{3}$

$=\sin \left(\frac{2\pi }{3}\right)$

$=\sin \pi -\frac{\pi }{3}$

$=\sin \frac{\pi }{3}$

$\sin \left(arc\cos (-\frac{1}{2})\right)=\frac{\sqrt{3}}{2}$

$\therefore \sin \left(arc\cos (-\frac{1}{2})\right)=\frac{\sqrt{3}}{2}$