Class 12: Inverse Trigonometric Functions

$\displaystyle \textbf{Question 1: }\text{Find the principal values of:}$
$\displaystyle \text{(i) }\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\qquad \text{(ii) }\sin^{-1}\left(\cos\frac{2\pi}{3}\right)$
$\displaystyle \text{(iii) }\sin^{-1}\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)\qquad \text{(iv) }\sin^{-1}\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)$
$\displaystyle \text{(v) }\sin^{-1}\left(\cos\frac{3\pi}{4}\right)\qquad \text{(vi) }\sin^{-1}\left(\tan\frac{5\pi}{4}\right)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The principal-value range of }\sin^{-1}x\text{ is }\left[-\frac{\pi}{2},\frac{\pi}{2}\right].$

$\displaystyle \text{(i) }\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)=\sin^{-1}\left[\sin\left(-\frac{\pi}{3}\right)\right]$
$\displaystyle \therefore \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)=-\frac{\pi}{3}$

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

$\displaystyle \text{(iii) }\frac{\sqrt{3}-1}{2\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}$
$\displaystyle =\sin\left(\frac{\pi}{4}-\frac{\pi}{6}\right)=\sin\frac{\pi}{12}$
$\displaystyle \therefore \sin^{-1}\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)=\frac{\pi}{12}$

$\displaystyle \text{(iv) }\frac{\sqrt{3}+1}{2\sqrt{2}}=\frac{\sqrt{6}+\sqrt{2}}{4}$
$\displaystyle =\sin\left(\frac{\pi}{4}+\frac{\pi}{6}\right)=\sin\frac{5\pi}{12}$
$\displaystyle \therefore \sin^{-1}\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)=\frac{5\pi}{12}$

$\displaystyle \text{(v) }\sin^{-1}\left(\cos\frac{3\pi}{4}\right)=\sin^{-1}\left(-\frac{\sqrt{2}}{2}\right)$
$\displaystyle =\sin^{-1}\left[\sin\left(-\frac{\pi}{4}\right)\right]=-\frac{\pi}{4}$

$\displaystyle \text{(vi) }\sin^{-1}\left(\tan\frac{5\pi}{4}\right)=\sin^{-1}(1)$
$\displaystyle =\sin^{-1}\left(\sin\frac{\pi}{2}\right)=\frac{\pi}{2}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }$
$\displaystyle \text{(i) }\sin^{-1}\left(\frac12\right)-2\sin^{-1}\left(\frac1{\sqrt2}\right)\qquad \text{(ii) }\sin^{-1}\left\{\cos\left(\sin^{-1}\frac{\sqrt3}{2}\right)\right\}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }\sin^{-1}\left(\frac12\right)-2\sin^{-1}\left(\frac1{\sqrt2}\right)$
$\displaystyle =\frac{\pi}{6}-2\left(\frac{\pi}{4}\right)$
$\displaystyle =\frac{\pi}{6}-\frac{\pi}{2}$
$\displaystyle =-\frac{\pi}{3}$

$\displaystyle \text{(ii) }\sin^{-1}\left\{\cos\left(\sin^{-1}\frac{\sqrt3}{2}\right)\right\}$
$\displaystyle =\sin^{-1}\left\{\cos\left(\sin^{-1}\left(\sin\frac{\pi}{3}\right)\right)\right\}$
$\displaystyle =\sin^{-1}\left(\cos\frac{\pi}{3}\right)$
$\displaystyle =\sin^{-1}\left(\frac12\right)$
$\displaystyle =\sin^{-1}\left(\sin\frac{\pi}{6}\right)$
$\displaystyle =\frac{\pi}{6}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Find the domain of each of the following functions:}$
$\displaystyle \text{(i) }f(x)=\sin^{-1}(x^2)$
$\displaystyle \text{(ii) }f(x)=\sin^{-1}x+\sin x$
$\displaystyle \text{(iii) }f(x)=\sin^{-1}\sqrt{x^2-1}$
$\displaystyle \text{(iv) }f(x)=\sin^{-1}x+\sin^{-1}(2x)$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }f(x)=\sin^{-1}(x^2)$
$\displaystyle \text{The domain of }\sin^{-1}y\text{ is }[-1,1].$
$\displaystyle \text{Hence }0\le x^2\le1.$
$\displaystyle \therefore -1\le x\le1.$
$\displaystyle \therefore \text{The domain is }[-1,1].$

$\displaystyle \text{(ii) }f(x)=\sin^{-1}x+\sin x$
$\displaystyle \text{The domain of }\sin^{-1}x\text{ is }[-1,1].$
$\displaystyle \text{The domain of }\sin x\text{ is }\mathbb{R}.$
$\displaystyle \therefore \text{The domain is }[-1,1].$

$\displaystyle \text{(iii) }f(x)=\sin^{-1}\sqrt{x^2-1}$
$\displaystyle \text{The domain of }\sin^{-1}y\text{ is }[-1,1].$
$\displaystyle \text{Also, }\sqrt{x^2-1}\ge0.$
$\displaystyle \therefore 0\le\sqrt{x^2-1}\le1.$
$\displaystyle \therefore 0\le x^2-1\le1.$
$\displaystyle \therefore 1\le x^2\le2.$
$\displaystyle \therefore x\in[-\sqrt2,-1]\cup[1,\sqrt2].$
$\displaystyle \therefore \text{The domain is }[-\sqrt2,-1]\cup[1,\sqrt2].$

$\displaystyle \text{(iv) }f(x)=\sin^{-1}x+\sin^{-1}(2x)$
$\displaystyle \text{The domain of }\sin^{-1}x\text{ is }[-1,1].$
$\displaystyle \text{The domain of }\sin^{-1}(2x)\text{ is }\left[-\frac12,\frac12\right].$
$\displaystyle \therefore \text{The domain is }\left[-\frac12,\frac12\right].$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{If }\sin^{-1}x+\sin^{-1}y+\sin^{-1}z+\sin^{-1}t=2\pi,\text{ find the value of}$
$\displaystyle x^2+y^2+z^2+t^2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The maximum value of each of }\sin^{-1}x,\sin^{-1}y,\sin^{-1}z\text{ and }\sin^{-1}t\text{ is }\frac{\pi}{2}.$
$\displaystyle \text{Therefore, their maximum possible sum is}$
$\displaystyle \frac{\pi}{2}+\frac{\pi}{2}+\frac{\pi}{2}+\frac{\pi}{2}=2\pi.$
$\displaystyle \text{Since the given sum is }2\pi,\text{ each term must attain its maximum value.}$
$\displaystyle \therefore \sin^{-1}x=\sin^{-1}y=\sin^{-1}z=\sin^{-1}t=\frac{\pi}{2}.$
$\displaystyle \therefore x=y=z=t=1.$
$\displaystyle \therefore x^2+y^2+z^2+t^2=1+1+1+1=4.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{If }(\sin^{-1}x)^2+(\sin^{-1}y)^2+(\sin^{-1}z)^2=\frac{3\pi^2}{4},\text{ find the value of}$
$\displaystyle x^2+y^2+z^2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }-\frac{\pi}{2}\le\sin^{-1}u\le\frac{\pi}{2},\text{ we have }(\sin^{-1}u)^2\le\left(\frac{\pi}{2}\right)^2.$
$\displaystyle \text{Therefore, }(\sin^{-1}x)^2+(\sin^{-1}y)^2+(\sin^{-1}z)^2\le3\left(\frac{\pi}{2}\right)^2=\frac{3\pi^2}{4}.$
$\displaystyle \text{Since the given sum equals }\frac{3\pi^2}{4},\text{ each term must be }\left(\frac{\pi}{2}\right)^2.$
$\displaystyle \therefore (\sin^{-1}x)^2=(\sin^{-1}y)^2=(\sin^{-1}z)^2=\left(\frac{\pi}{2}\right)^2.$
$\displaystyle \therefore \sin^{-1}x=\pm\frac{\pi}{2},\ \sin^{-1}y=\pm\frac{\pi}{2},\ \sin^{-1}z=\pm\frac{\pi}{2}.$
$\displaystyle \therefore x=\pm1,\ y=\pm1,\ z=\pm1.$
$\displaystyle \therefore x^2+y^2+z^2=1+1+1=3.$
$\displaystyle \\$