Class 12: Inverse Trigonometric Functions

$\displaystyle \textbf{Question 1: }\text{Evaluate: }\cos\left(\sin^{-1}\frac35+\sin^{-1}\frac5{13}\right)$
$\displaystyle \text{Answer:}$
$\displaystyle \cos\left(\sin^{-1}\frac35+\sin^{-1}\frac5{13}\right)$
$\displaystyle =\cos\left\{\sin^{-1}\left[\frac35\sqrt{1-\left(\frac5{13}\right)^2}+\frac5{13}\sqrt{1-\left(\frac35\right)^2}\right]\right\}$
$\displaystyle =\cos\left\{\sin^{-1}\left(\frac35\cdot\frac{12}{13}+\frac5{13}\cdot\frac45\right)\right\}$
$\displaystyle =\cos\left\{\sin^{-1}\left(\frac{36}{65}+\frac{20}{65}\right)\right\}$
$\displaystyle =\cos\left(\sin^{-1}\frac{56}{65}\right)$
$\displaystyle =\sqrt{1-\left(\frac{56}{65}\right)^2}$
$\displaystyle =\frac{33}{65}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Prove the following results:}$
$\displaystyle \text{(i) }\sin^{-1}\left(\frac{63}{65}\right)=\sin^{-1}\left(\frac5{13}\right)+\cos^{-1}\left(\frac35\right)\qquad\text{[CBSE 2012]}$
$\displaystyle \text{(ii) }\sin^{-1}\left(\frac5{13}\right)+\cos^{-1}\left(\frac35\right)=\tan^{-1}\left(\frac{63}{16}\right)$
$\displaystyle \text{(iii) }\frac{9\pi}{8}-\frac94\sin^{-1}\left(\frac13\right)=\frac94\sin^{-1}\left(\frac{2\sqrt2}{3}\right)$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i)}$
$\displaystyle \text{RHS}=\sin^{-1}\left(\frac5{13}\right)+\cos^{-1}\left(\frac35\right)$
$\displaystyle =\sin^{-1}\left(\frac5{13}\right)+\sin^{-1}\left(\frac45\right)$
$\displaystyle \qquad\left[\because\ \cos^{-1}x=\sin^{-1}\sqrt{1-x^2},\ 0\leq x\leq1\right]$
$\displaystyle \text{Let }\alpha=\sin^{-1}\left(\frac5{13}\right)\text{ and }\beta=\sin^{-1}\left(\frac45\right).$
$\displaystyle \therefore\sin\alpha=\frac5{13},\quad\cos\alpha=\frac{12}{13},\quad\sin\beta=\frac45,\quad\cos\beta=\frac35.$
$\displaystyle \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$
$\displaystyle =\frac5{13}\cdot\frac35+\frac{12}{13}\cdot\frac45$
$\displaystyle =\frac{15}{65}+\frac{48}{65}=\frac{63}{65}.$
$\displaystyle \text{Also, }\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$
$\displaystyle =\frac{12}{13}\cdot\frac35-\frac5{13}\cdot\frac45$
$\displaystyle =\frac{36}{65}-\frac{20}{65}=\frac{16}{65}>0.$
$\displaystyle \text{Hence, }0<\alpha+\beta<\frac{\pi}{2}.$
$\displaystyle \therefore\alpha+\beta=\sin^{-1}\left(\frac{63}{65}\right)$
$\displaystyle \Rightarrow\sin^{-1}\left(\frac5{13}\right)+\cos^{-1}\left(\frac35\right)=\sin^{-1}\left(\frac{63}{65}\right)=\text{LHS}$
$\displaystyle \\$

$\displaystyle \text{(ii)}$
$\displaystyle \text{LHS}=\sin^{-1}\left(\frac5{13}\right)+\cos^{-1}\left(\frac35\right)$
$\displaystyle =\sin^{-1}\left(\frac5{13}\right)+\sin^{-1}\left(\frac45\right)$
$\displaystyle =\sin^{-1}\left(\frac{63}{65}\right)$
$\displaystyle \qquad\left[\text{From part (i)}\right]$
$\displaystyle =\tan^{-1}\left[\frac{\frac{63}{65}}{\sqrt{1-\left(\frac{63}{65}\right)^2}}\right]$
$\displaystyle \qquad\left[\because\ \sin^{-1}x=\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right),\ 0\leq x<1\right]$
$\displaystyle =\tan^{-1}\left[\frac{\frac{63}{65}}{\sqrt{\frac{4225-3969}{4225}}}\right]$
$\displaystyle =\tan^{-1}\left(\frac{\frac{63}{65}}{\frac{16}{65}}\right)$
$\displaystyle =\tan^{-1}\left(\frac{63}{16}\right)=\text{RHS}$
$\displaystyle \\$

$\displaystyle \text{(iii)}$
$\displaystyle \text{LHS}=\frac{9\pi}{8}-\frac94\sin^{-1}\left(\frac13\right)$
$\displaystyle =\frac94\left[\frac{\pi}{2}-\sin^{-1}\left(\frac13\right)\right]$
$\displaystyle =\frac94\cos^{-1}\left(\frac13\right)$
$\displaystyle =\frac94\sin^{-1}\sqrt{1-\left(\frac13\right)^2}$
$\displaystyle =\frac94\sin^{-1}\sqrt{\frac89}$
$\displaystyle =\frac94\sin^{-1}\left(\frac{2\sqrt2}{3}\right)$
$\displaystyle =\text{RHS}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Solve the following:}$
$\displaystyle \text{(i) }\sin^{-1}x+\sin^{-1}2x=\frac{\pi}{3}$
$\displaystyle \text{(ii) }\cos^{-1}x+\sin^{-1}\frac{x}{2}-\frac{\pi}{6}=0\qquad\text{[CBSE 2012]}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i)}$
$\displaystyle \sin^{-1}x+\sin^{-1}2x=\frac{\pi}{3}$
$\displaystyle \Rightarrow\sin^{-1}x=\frac{\pi}{3}-\sin^{-1}2x$
$\displaystyle \text{Taking sine on both sides,}$
$\displaystyle x=\sin\left(\frac{\pi}{3}-\sin^{-1}2x\right)$
$\displaystyle \Rightarrow x=\sin\frac{\pi}{3}\cos(\sin^{-1}2x)-\cos\frac{\pi}{3}\sin(\sin^{-1}2x)$
$\displaystyle \Rightarrow x=\frac{\sqrt3}{2}\sqrt{1-4x^2}-\frac12(2x)$
$\displaystyle \Rightarrow2x=\frac{\sqrt3}{2}\sqrt{1-4x^2}$
$\displaystyle \Rightarrow4x=\sqrt3\sqrt{1-4x^2}$
$\displaystyle \text{Since the right-hand side is non-negative, }x\geq0.$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle 16x^2=3(1-4x^2)$
$\displaystyle \Rightarrow16x^2=3-12x^2$
$\displaystyle \Rightarrow28x^2=3$
$\displaystyle \Rightarrow x^2=\frac3{28}$
$\displaystyle \Rightarrow x=\frac12\sqrt{\frac37}$
$\displaystyle \therefore x=\frac12\sqrt{\frac37}$
$\displaystyle \\$

$\displaystyle \text{(ii)}$
$\displaystyle \cos^{-1}x+\sin^{-1}\frac{x}{2}-\frac{\pi}{6}=0$
$\displaystyle \Rightarrow\cos^{-1}x+\sin^{-1}\frac{x}{2}=\frac{\pi}{6}$
$\displaystyle \Rightarrow\frac{\pi}{2}-\sin^{-1}x+\sin^{-1}\frac{x}{2}=\frac{\pi}{6}$
$\displaystyle \Rightarrow\sin^{-1}x-\sin^{-1}\frac{x}{2}=\frac{\pi}{3}$
$\displaystyle \Rightarrow\sin^{-1}x=\frac{\pi}{3}+\sin^{-1}\frac{x}{2}$
$\displaystyle \text{Taking sine on both sides,}$
$\displaystyle x=\sin\left(\frac{\pi}{3}+\sin^{-1}\frac{x}{2}\right)$
$\displaystyle \Rightarrow x=\frac{\sqrt3}{2}\sqrt{1-\frac{x^2}{4}}+\frac12\cdot\frac{x}{2}$
$\displaystyle \Rightarrow x=\frac{\sqrt3}{4}\sqrt{4-x^2}+\frac{x}{4}$
$\displaystyle \Rightarrow\frac{3x}{4}=\frac{\sqrt3}{4}\sqrt{4-x^2}$
$\displaystyle \Rightarrow3x=\sqrt3\sqrt{4-x^2}$
$\displaystyle \text{Since the right-hand side is non-negative, }x\geq0.$
$\displaystyle \text{Squaring both sides,}$
$\displaystyle 9x^2=3(4-x^2)$
$\displaystyle \Rightarrow9x^2=12-3x^2$
$\displaystyle \Rightarrow12x^2=12$
$\displaystyle \Rightarrow x^2=1$
$\displaystyle \text{Since }x\geq0,\ x=1.$
$\displaystyle \text{Also, }\cos^{-1}1+\sin^{-1}\frac12=0+\frac{\pi}{6}=\frac{\pi}{6}.$
$\displaystyle \therefore x=1$
$\displaystyle \\$