Question 1: Find the domain of definition $f(x) = \cos^{-1}(x^2 -4).$

$\text{The domain of } \cos^{-1} x \text{ is } [-1, 1].$

$\text{So, the domain of } \cos^{-1} (x^2-4) \text{ is a set of all values of } x \text{ satisfying }$

$-1 \leq x^2 - 4 \leq 1$

$\Rightarrow 3 \leq x^2 \leq 5$

$\Rightarrow x \in [ -\sqrt{5}, - \sqrt{3} ] \cup [ \sqrt{3}, \sqrt{5} ]$

$\text{Hence, the domain of } \cos^{-1}(x^2 -4) \text{ is } [ -\sqrt{5}, - \sqrt{3} ] \cup [ \sqrt{3}, \sqrt{5} ]$

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Question 2: Find the domain of $f(x) = 2 \cos^{-1} 2x + \sin^{-1}.$

$\text{The domain of } \cos^{-1} x \text{ is } [-1, 1].$

$\text{So, the domain of } \cos^{-1} 2x \text{ is a set of all values of } x \text{ satisfying }$

$\displaystyle -1 \leq 2x \leq 1 \Rightarrow -\frac{1}{2} \leq x \leq \frac{1}{2}$

$\text{The domain of } \sin^{-1} x \text{ is } [-1, 1].$

$\displaystyle \text{Hence, the domain of } f(x) = \Bigg[ -\frac{1}{2}, \frac{1}{2} \Bigg] \cap [-1. 1] = \Bigg[ -\frac{1}{2}, \frac{1}{2} \Bigg]$

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Question 3: Find the domain of $f(x) = \cos^{-1} x + \cos x.$

$\text{The domain of } \cos^{-1} x \text{ is } [-1, 1].$

$\text{The domain of } \cos x \text{ is } [-\infty, \infty].$

$\displaystyle \text{Hence, the domain of } f(x) = [-1, 1] \cap [-\infty, \infty] = [-1, 1]$

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Question 4: Find the principal value of each of the following:

$\displaystyle \text{(i) } \cos^{-1} \Bigg( -\frac{\sqrt{3} }{2} \Bigg)$

$\displaystyle \text{(ii) } \cos^{-1} \Bigg( -\frac{1 }{\sqrt{2}} \Bigg)$

$\displaystyle \text{(iii) } \cos^{-1} \Bigg( \sin \frac{4\pi }{3} \Bigg)$

$\displaystyle \text{(iv) } \cos^{-1} \Bigg( \tan \frac{3\pi }{4} \Bigg)$

$\text{For any } x \in [-1, 1 ], \cos^{-1} x \text{ represents an angle in } [0, \pi ] \text{ whose cosine is } x. \\ \\ \text{Therefore, }$

$\displaystyle \text{(i) } \cos^{-1} \Bigg( -\frac{\sqrt{3} }{2} \Bigg) \\ \\ = \Bigg( \text{An angle } \theta \in [ 0, \pi ] \text{ such that } \cos \theta = -\frac{\sqrt{3} }{2} = \cos \Bigg( \frac{5\pi}{6} \Bigg) \Bigg) = \frac{5\pi}{6}$

$\displaystyle \text{(ii) } \cos^{-1} \Bigg( -\frac{1 }{\sqrt{2}} \Bigg) \\ \\ = \Bigg( \text{An angle } \theta \in [ 0, \pi ] \text{ such that } \cos \theta = -\frac{1 }{\sqrt{2}} = \cos \Bigg( \frac{3\pi}{4} \Bigg) \Bigg) = \frac{3\pi}{4}$

$\text{(iii) } \cos^{-1} \Bigg( \sin \frac{4\pi }{3} \Bigg) \\ \\ = \Bigg( \text{An angle } \theta \in [ 0, \pi ] \text{ such that } \cos \theta = \sin \frac{4\pi }{3} = -\frac{\sqrt{3} }{2} = \cos \Bigg( \frac{5\pi}{6} \Bigg) \Bigg) = \frac{5\pi}{6}$

$\displaystyle \text{(iv) } \cos^{-1} \Bigg( \tan \frac{3\pi }{4} \Bigg) \\ \\ = \Bigg( \text{An angle } \theta \in [ 0, \pi ] \text{ such that } \cos \theta = \tan \frac{3\pi }{4} = -1 = \cos (\pi) \Bigg) = \pi$

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Question 5: Find the principal values, evaluate each of the following:

$\displaystyle \text{(i) } \cos^{-1} \Bigg( \frac{1 }{2} \Bigg) + 2\sin^{-1} \Bigg( \frac{1 }{2} \Bigg)$

$\displaystyle \text{(ii) } \cos^{-1} \Bigg( \frac{1 }{2} \Bigg) - 2\sin^{-1} \Bigg( -\frac{1 }{2} \Bigg) \hspace{2.0cm} \textbf{ CBSE 2012}$

$\displaystyle \text{(iii) } \sin^{-1} \Bigg( - \frac{1 }{2} \Bigg) + 2\cos^{-1} \Bigg( - \frac{\sqrt{3} }{2} \Bigg)$

$\displaystyle \text{(iv) } \sin^{-1} \Bigg( - \frac{\sqrt{3} }{2} \Bigg) + \cos^{-1} \Bigg( \frac{\sqrt{3} }{2} \Bigg)$

$\displaystyle \text{(i) } \cos^{-1} \Bigg( \frac{1 }{2} \Bigg) + 2\sin^{-1} \Bigg( \frac{1 }{2} \Bigg) \\ \\ \\ = \cos^{-1} \Bigg( \cos \frac{\pi}{3} \Bigg) + 2\sin^{-1} \Bigg( \sin \frac{\pi}{6} \Bigg) \\ \\ \\ = \frac{\pi}{3}+ 2 \Bigg( \frac{\pi}{6} \Bigg) \\ \\ \\ = \frac{2\pi}{3}$
$\displaystyle \text{(ii) } \cos^{-1} \Bigg( \frac{1 }{2} \Bigg) - 2\sin^{-1} \Bigg( -\frac{1 }{2} \Bigg) \hspace{2.0cm} \\ \\ \\ = \cos^{-1} \Bigg( \cos \frac{\pi}{3} \Bigg) - 2\sin^{-1} \Bigg( -\sin \frac{\pi}{6} \Bigg) \\ \\ \\ = \frac{\pi}{3}- 2 \Bigg( \frac{-\pi}{6} \Bigg) \\ \\ \\ = \frac{2\pi}{3}$
$\displaystyle \text{(iii) } \sin^{-1} \Bigg( - \frac{1 }{2} \Bigg) + 2\cos^{-1} \Bigg( - \frac{\sqrt{3} }{2} \Bigg) \\ \\ \\ = \sin^{-1} \Bigg[ \sin \Bigg( -\frac{\pi}{6} \Bigg) \Bigg) + 2\cos^{-1} \Bigg( \cos \Bigg(\frac{5\pi}{6} \Bigg) \Bigg] \\ \\ = -\frac{\pi}{6} + 2 \Bigg( \frac{5\pi}{6} \Bigg) \\ \\ \\ = \frac{9\pi}{6} \\ \\ \\ = \frac{3\pi}{2}$
$\displaystyle \text{(iv) } \sin^{-1} \Bigg( - \frac{\sqrt{3} }{2} \Bigg) + \cos^{-1} \Bigg( \frac{\sqrt{3} }{2} \Bigg) \\ \\ \\ = \sin^{-1} \Bigg[ \sin \Bigg( -\frac{\pi}{2} \Bigg) \Bigg) + \cos^{-1} \Bigg( \cos \Bigg(\frac{\pi}{6} \Bigg) \Bigg] \\ \\ \\ = -\frac{\pi}{3} + \frac{\pi}{6} \\ \\ \\ -\frac{\pi}{6}$