Find the value of the following:$cos^{-1} \big(- \frac {1} {2} \big)+2 sin^{-1}\frac {1} {2}$

<div class="clay6-toolbox"><b>Toolbox:</b><ul><li class="clay6-basic" id="pr00">The range of the principal value of $\sin^{-1}x$ is $\left [ -\large\frac{\pi}{2}, \large\frac{\pi}{2} \right ]$</li><li class="clay6-basic" id="pr01">The range of the principal value of $\cos^{-1}x$ is $\left [ 0,\pi \right ]$</li><li class="clay6-basic" id="pr02">In such problems, we find the principal values of each of the inverse trignometric identities and arrive at the final answer.</li></ul></div><div class="clay6-step-odd"><div class="clay6-basic" id="pr10">Let $\sin^{-1}(\large\frac{1}{2}) = x$ $\Rightarrow \sin x = \large\frac{1}{2}$</div><div class="clay6-basic" id="pr11">$\Rightarrow \sin x = \large\frac{1}{2} = \sin \large\frac{\pi}{6}$</div><div class="clay6-basic" id="pr12">$$\therefore$$ $x =\large\frac{\pi}{6} \; \in \; \left [ -\large\frac{\pi}{2}, \large\frac{\pi}{2} \right ]$</div><div class="clay6-basic" id="pr13">Let $\cos^{-1}(-\large\frac{1}{2}) = y\:$ $\Rightarrow\: \cos y =- \large\frac{1}{2}$</div><div class="clay6-basic" id="pr14">$\Rightarrow \cos y = -\large\frac{1}{2} = \cos(\pi- \large\frac{\pi}{3})$$=cos \large\frac{2\pi}{3}</div><div class="clay6-basic" id="pr15">$$\therefore$$ y = \large\frac{2\pi}{3}$$ \; \in \; \left [ 0,\pi \right ]$</div><div class="clay6-basic" id="pr16">$\cos^{-1}(-\large\frac{1}{2}) + 2 \sin^{-1}(\large\frac{1}{2}) = y + 2x$</div><div class="clay6-basic" id="pr17">$\Rightarrow \large\frac{2\pi}{3} $$+ 2\large\frac{\pi}{6} = \large\frac{2\pi}{3} + \large\frac{\pi}{3}$$ = \pi$</div></div>
edited Mar 20, 2014