# Evaluate : $\cos\; \large \bigg( \frac{\pi}{2} -$$\cos^{-1}\; \large\bigg (-\frac{\sqrt 3}{2}\bigg)\bigg) (A)\; \large\frac{1}{2} \\ (B)\; -\large\frac{1}{2} \\ (C)\; 1 \\ (D)\;\large\frac{3}{2} ## 1 Answer Toolbox: • \cos^{-1} (-x) = \pi - \cos^{-1}x • Principal interval of cos is $$[0,\pi]$$ • (\cos\large\frac{\pi}{6}=\large\frac{\sqrt{3}}{2}) \rightarrow \cos^{-1} \large (\frac{\sqrt 3}{2})$$= \large \frac{\pi}{6}$
• $\cos\; - \large (\frac{\pi}{3}) $$= \frac{1}{2} Given: \cos \large ( \frac{\pi}{2} -$$ \cos^{-1} \large (-\frac{\sqrt 3}{2}))$
We know that $\cos^{-1} (-x) = \pi - \cos^{-1}x$
$\Rightarrow$ $\cos \large ( \frac{\pi}{2} -$$\cos^{-1} \large (-\frac{\sqrt 3}{2}))\;$$= \cos \large ( \frac{\pi}{2} -$$\pi + \cos^{-1} \large (\frac{\sqrt 3}{2})) = \cos \large ($$\cos^{-1} \large (\frac{\sqrt 3}{2}) - \frac{\pi}{2})$
We know that $\cos^{-1} \large (\frac{\sqrt 3}{2}) $$= \large \frac{\pi}{6} \Rightarrow \cos \large ( \frac{\pi}{2} -$$ \cos^{-1} \large (-\frac{\sqrt 3}{2}))\;$ $= \cos \large (\frac{\pi}{6} - \frac{\pi}{2})$$= \cos - \large (\frac{\pi}{3}) We know that: \cos - \large (\frac{\pi}{3})$$= \frac{1}{2}$