\displaystyle \textbf{Question 1: }\text{Find the domain of definition }f(x)=\cos^{-1}(x^2-4).
\displaystyle \text{Answer:}
\displaystyle \text{The domain of }\cos^{-1}x\text{ is }[-1,1].
\displaystyle \text{Hence, }-1\le x^2-4\le1.
\displaystyle \therefore 3\le x^2\le5.
\displaystyle \therefore x\in[-\sqrt5,-\sqrt3]\cup[\sqrt3,\sqrt5].
\displaystyle \therefore \text{The domain of }f(x)\text{ is }[-\sqrt5,-\sqrt3]\cup[\sqrt3,\sqrt5].
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the domain of }f(x)=2\cos^{-1}(2x)+\sin^{-1}x.
\displaystyle \text{Answer:}
\displaystyle \text{The domain of }\cos^{-1}x\text{ is }[-1,1].
\displaystyle \text{Hence, }-1\le2x\le1.
\displaystyle \therefore -\frac12\le x\le\frac12.
\displaystyle \text{The domain of }\sin^{-1}x\text{ is }[-1,1].
\displaystyle \therefore \text{The required domain is }\left[-\frac12,\frac12\right]\cap[-1,1].
\displaystyle =\left[-\frac12,\frac12\right].
\displaystyle \therefore \text{The domain of }f(x)\text{ is }\left[-\frac12,\frac12\right].
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the domain of }f(x)=\cos^{-1}x+\cos x.
\displaystyle \text{Answer:}
\displaystyle \text{The domain of }\cos^{-1}x\text{ is }[-1,1].
\displaystyle \text{The domain of }\cos x\text{ is }\mathbb{R}.
\displaystyle \therefore \text{The required domain is }[-1,1]\cap\mathbb{R}.
\displaystyle =[-1,1].
\displaystyle \therefore \text{The domain of }f(x)\text{ is }[-1,1].
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the principal value of each of the following:}
\displaystyle \text{(i) }\cos^{-1}\left(-\frac{\sqrt3}{2}\right)\qquad \text{(ii) }\cos^{-1}\left(-\frac{1}{\sqrt2}\right)
\displaystyle \text{(iii) }\cos^{-1}\left(\sin\frac{4\pi}{3}\right)\qquad \text{(iv) }\cos^{-1}\left(\tan\frac{3\pi}{4}\right)
\displaystyle \text{Answer:}
\displaystyle \text{For any }x\in[-1,1],\ \cos^{-1}x\text{ represents an angle in }[0,\pi]\text{ whose cosine is }x.
\displaystyle \text{(i) }\cos^{-1}\left(-\frac{\sqrt3}{2}\right)=\cos^{-1}\left(\cos\frac{5\pi}{6}\right)=\frac{5\pi}{6}.
\displaystyle \text{(ii) }\cos^{-1}\left(-\frac{1}{\sqrt2}\right)=\cos^{-1}\left(\cos\frac{3\pi}{4}\right)=\frac{3\pi}{4}.
\displaystyle \text{(iii) }\cos^{-1}\left(\sin\frac{4\pi}{3}\right)=\cos^{-1}\left(-\frac{\sqrt3}{2}\right)
\displaystyle =\cos^{-1}\left(\cos\frac{5\pi}{6}\right)=\frac{5\pi}{6}.
\displaystyle \text{(iv) }\cos^{-1}\left(\tan\frac{3\pi}{4}\right)=\cos^{-1}(-1)
\displaystyle =\cos^{-1}(\cos\pi)=\pi.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the principal values and evaluate each of the following:}
\displaystyle \text{(i) }\cos^{-1}\left(\frac12\right)+2\sin^{-1}\left(\frac12\right)
\displaystyle \text{(ii) }\cos^{-1}\left(\frac12\right)-2\sin^{-1}\left(-\frac12\right)\hfill\text{[CBSE 2012]}
\displaystyle \text{(iii) }\sin^{-1}\left(-\frac12\right)+2\cos^{-1}\left(-\frac{\sqrt3}{2}\right)
\displaystyle \text{(iv) }\sin^{-1}\left(-\frac{\sqrt3}{2}\right)+\cos^{-1}\left(\frac{\sqrt3}{2}\right)
\displaystyle \text{Answer:}

\displaystyle \text{(i) }\cos^{-1}\left(\frac12\right)+2\sin^{-1}\left(\frac12\right)
\displaystyle =\cos^{-1}\left(\cos\frac{\pi}{3}\right)+2\sin^{-1}\left(\sin\frac{\pi}{6}\right)
\displaystyle =\frac{\pi}{3}+2\left(\frac{\pi}{6}\right)
\displaystyle =\frac{2\pi}{3}

\displaystyle \text{(ii) }\cos^{-1}\left(\frac12\right)-2\sin^{-1}\left(-\frac12\right)
\displaystyle =\cos^{-1}\left(\cos\frac{\pi}{3}\right)-2\sin^{-1}\left(\sin\left(-\frac{\pi}{6}\right)\right)
\displaystyle =\frac{\pi}{3}-2\left(-\frac{\pi}{6}\right)
\displaystyle =\frac{2\pi}{3}

\displaystyle \text{(iii) }\sin^{-1}\left(-\frac12\right)+2\cos^{-1}\left(-\frac{\sqrt3}{2}\right)
\displaystyle =\sin^{-1}\left(\sin\left(-\frac{\pi}{6}\right)\right)+2\cos^{-1}\left(\cos\frac{5\pi}{6}\right)
\displaystyle =-\frac{\pi}{6}+2\left(\frac{5\pi}{6}\right)
\displaystyle =\frac{9\pi}{6}=\frac{3\pi}{2}

\displaystyle \text{(iv) }\sin^{-1}\left(-\frac{\sqrt3}{2}\right)+\cos^{-1}\left(\frac{\sqrt3}{2}\right)
\displaystyle =\sin^{-1}\left(\sin\left(-\frac{\pi}{3}\right)\right)+\cos^{-1}\left(\cos\frac{\pi}{6}\right)
\displaystyle =-\frac{\pi}{3}+\frac{\pi}{6}
\displaystyle =-\frac{\pi}{6}
\displaystyle \\

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