Class 12: Inverse Trigonometric Functions – Exercise 4.1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Find the principal values of:

$\displaystyle \text{(i) } \sin^{-1} \Bigg( -\frac{\sqrt{3}}{2} \Bigg) \hspace{1.0cm} \text{(ii) } \sin^{-1} \Bigg( \cos \frac{2\pi}{3} \Bigg) \hspace{1.0cm} \text{(iii) } \sin^{-1} \Bigg( \frac{\sqrt{3}-1}{2\sqrt{2}} \Bigg)$

$\displaystyle \text{(iv) } \sin^{-1} \Bigg( \frac{\sqrt{3}+1}{2\sqrt{2}} \Bigg) \hspace{1.0cm} \text{(v) } \sin^{-1} \Bigg( \cos \frac{3\pi}{4} \Bigg) \hspace{1.0cm} \text{(vi) } \sin^{-1} \Bigg( \tan \frac{5\pi}{4} \Bigg)$

Answer:

$\displaystyle \text{(i) } \sin^{-1} \Big( -\frac{\sqrt{3}}{2} \Big) = \sin^{-1} \Big[ \sin \Big( -\frac{\pi}{3} \Big) \Big] = -\frac{\pi}{3} \\ \\ \\ \text{(ii) } \sin^{-1} \Big( \cos \frac{2\pi}{3} \Big) = \sin^{-1} \Big( -\frac{1}{2} \Big) = \sin^{-1} \Big[ \sin \Big( -\frac{\pi}{6} \Big) \Big] = -\frac{\pi}{6} \\ \\ \\ \text{(iii) } \sin^{-1} \Big( \frac{\sqrt{3}-1}{2\sqrt{2}} \Big) = \sin^{-1} \Big( \sin \frac{\pi}{12} \Big) = \frac{\pi}{12} \\ \\ \\ \text{(iv) } \sin^{-1} \Big( \frac{\sqrt{3}+1}{2\sqrt{2}} \Big) = \sin^{-1} \Big( \sin \frac{5\pi}{12} \Big) = \frac{5\pi}{12} \\ \\ \\ \text{(v) } \sin^{-1} \Big( \cos \frac{3\pi}{4} \Big) = \sin^{-1} \Big( - \frac{\sqrt{2}}{2} \Big) = \sin^{-1} \Big[ \sin \Big( -\frac{\pi}{4} \Big) \Big] = -\frac{\pi}{4} \\ \\ \\ \text{(vi) } \sin^{-1} \Big( \tan \frac{5\pi}{4} \Big) = \sin^{-1} (1) = \sin^{-1} \Big[ \sin \Big( \frac{\pi}{2} \Big) \Big] = \frac{\pi}{2}$

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Question 2:

$\displaystyle \text{(i) } \sin^{-1} \Big( \frac{1}{2} \Big) - 2 \sin^{-1} \Big( \frac{1}{\sqrt{2}} \Big) \hspace{1.0cm} \text{(ii) } \sin^{-1} \Big\{ \cos \Big( \sin^{-1} \frac{\sqrt{3}}{2} \Big) \Big\}$

Answer:

$\displaystyle \text{(i) } \sin^{-1} \Bigg( \frac{1}{2} \Bigg) - 2 \sin^{-1} \Bigg( \frac{1}{\sqrt{2}} \Bigg) = \sin^{-1} \frac{1}{2} - \sin^{-1} 2 \times \frac{1}{\sqrt{2}} \sqrt{1 - \Bigg( \frac{1}{\sqrt{2}} \Bigg)^2} \\ \\ \\ = \sin^{-1} \frac{1}{2} - \sin^{-1}\sqrt{2} \times \frac{1}{\sqrt{2}} \\ \\ \\ = \sin^{-1} \frac{1}{2} - \sin^{-1} 1 \\ \\ \\ = \sin^{-1} \Big( \sin \frac{\pi}{6} \Big) - \sin^{-1} \Big( \sin \frac{\pi}{2} \Big) \\ \\ \\ = \frac{\pi}{6} - \frac{\pi}{2} \\ \\ \\ = - \frac{\pi}{3}$

$\displaystyle \text{(ii) } \sin^{-1} \Big\{ \cos \Big( \sin^{-1} \frac{\sqrt{3}}{2} \Big) \Big\} = \sin^{-1} \Big\{ \cos \Big( \sin^{-1} \sin \frac{\pi}{3} \Big) \Big\} \\ \\ \\ = \sin^{-1} \Big\{ \cos \Big( \frac{\pi}{3} \Big) \Big\} \\ \\ \\ = \sin^{-1} \Big\{ \frac{1}{2} \Big\} \\ \\ \\ = \sin^{-1} \Big\{ \sin \frac{\pi}{6} \Big\} \\ \\ \\ = \frac{\pi}{6}$

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Question 3: Find the domain of each of the following functions:

$\text{(i) } f(x) = \sin^{-1} x^2$

$\text{(ii) } f(x) = \sin^{-1} x + \sin x$

$\text{(iii) } f(x) = \sin^{-1} \sqrt{x^2-1}$

$\text{(iv) } f(x) = \sin^{-1} x + \sin^{-1} 2x$

Answer:

$\text{(i) } f(x) = \sin^{-1} x^2$

$\text{To the domain of } \sin^{-1} \text{ which is } [-1, 1 ]$

$\text{Therefore } x^2 \in [0, 1] as x^2 \text{ cannot be negative }$

$\text{Therefore } x \in [-1, 1]$

$\text{Hence, the domain is } [-1, 1]$

$\text{(ii) } f(x) = \sin^{-1} x + \sin x$

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

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

$\text{Therefore the intersection of } \sin^{-1} x \text{ and } \sin x\text{ is } [-1, 1]$

$\text{Hence, the domain is } [-1, 1]$

$\text{(iii) } f(x) = \sin^{-1} \sqrt{x^2-1}$

$\text{To the domain of } \sin^{-1} y \text{ which is } [-1, 1]$

$\text{Therefore } x^2 -1 \in [0, 1] \text{ as square root can not be negative. }$

$\Rightarrow x^2 \in [0, 1]$

$\Rightarrow x \in [-\sqrt{2}, -1 ] \cup [1, \sqrt{2} ]$

$\text{Hence the domain is } [-\sqrt{2}, -1 ] \cup [1, \sqrt{2} ]$

$\text{(iv) } f(x) = \sin^{-1} x + \sin^{-1} 2x$

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

$\displaystyle \text{The domain of } \sin^{-1} 2x \text{ is} \Big[- \frac{1}{2}, \frac{1}{2} \Big]$

$\displaystyle \text{Therefore the intersection of } \sin^{-1} x \text{ and } \sin^{-1} 2x \text{ is } \Big[- \frac{1}{2}, \frac{1}{2} \Big]$

$\displaystyle \text{Hence, the domain is } \Big[- \frac{1}{2}, \frac{1}{2} \Big]$

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$\text{Question 4: If } \sin^{-1} x + \sin^{-1} y + \sin^{-1} z + \sin^{-1} t = 2\pi, \text{ then find the value of } \\ \\ x^2 + y^2 + z^2 + t^2$

Answer:

$\displaystyle \text{We know that the maximum value of } \sin^{-1} x, \sin^{-1} y, \sin^{-1} z \text{ and } \sin^{-1} t \text{ is } \frac{\pi}{2}$

$\displaystyle \text{Now, } \\ \\ \text{ LHS } = \sin^{-1} x + \sin^{-1} y + \sin^{-1} z + \sin^{-1} t = \frac{\pi}{2} +\frac{\pi}{2} +\frac{\pi}{2} +\frac{\pi}{2} = 2\pi = \text{ RHS }$

$\displaystyle \text{Now, } \\ \\ \sin^{-1} x = \frac{\pi}{2}, \ \sin^{-1} y = \frac{\pi}{2}, \ \sin^{-1} z = \frac{\pi}{2}, \text{ and } \sin^{-1} t = \frac{\pi}{2},$

$\displaystyle \Rightarrow x = 1, y = 1, z = 1 \text{ and } t = 1$

$\displaystyle \text{Therefore } x^2 + y^2 + z^2 + t^2 = 1 + 1 + 1 + 1 = 4$

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$\text{Question 5: If } (\sin^{-1} x)^2+ (\sin^{-1} y)^2+ (\sin^{-1} z)^2 = \frac{3}{4} \pi^2, \text{ then find the value of } \\ \\ x^2 + y^2 + z^2$

Answer:

$\displaystyle \text{We know that the maximum value of } \sin^{-1} x, \sin^{-1} y, \sin^{-1} z \text{ is } \frac{\pi}{2} \\ \\ \text{ and the minimum value of } \sin^{-1} x, \sin^{-1} y, \sin^{-1} z \text{ is } -\frac{\pi}{2}$

For maximum value

$\displaystyle \text{ LHS } = (\sin^{-1} x)^2+ (\sin^{-1} y)^2 + (\sin^{-1} z)^2 \\ \\ = \Big( \frac{\pi}{2} \Big)^2 + \Big( \frac{\pi}{2} \Big)^2 + \Big( \frac{\pi}{2} \Big)^2 = \frac{3}{4} \pi^2 = \text{ RHS }$

For minimum value

$\displaystyle \text{ LHS } = (\sin^{-1} x)^2+ (\sin^{-1} y)^2 + (\sin^{-1} z)^2 \\ \\ = \Big(- \frac{\pi}{2} \Big)^2 + \Big( -\frac{\pi}{2} \Big)^2 + \Big( -\frac{\pi}{2} \Big)^2 = \frac{3}{4} \pi^2 = \text{ RHS }$