Find ${\displaystyle -1<x<0}$ such that

${\displaystyle \sin \left({\cos }^{-1}x\right)=\frac{1}{2}.}$

Solution:

sin(cos$^{-1}$(x) = $\frac{1}{2}$ = sin($\frac{\pi}{6}$)

${\displaystyle {\cos }^{-1}x=\frac{\pi }{6}}$

$x=\cos\frac{\pi}{6}=\frac{\sqrt{3}}{6}$

But, x must be negative.

If -1 < x < 0 , then

cos$^{-1}x \in ({\frac{\pi}{2}},{\pi})$ But sin$^{-1} ({\frac{1}{2}})$ $\in ({\frac{-\pi}{2},\frac{\pi}{2}}) \leftarrow$ Range set of sin$^{-1}$

Find

${\displaystyle -1<x<0}$

such that

$\sin\left(\cos^{-1}x\right)=\frac{1}{2}=\sin\frac{\pi}{6}$

${\displaystyle {\cos }^{-1}x=\frac{\pi }{6}={\sin }^{-1}\frac{1}{2}}$

$x=\cos\frac{\pi}{6}=\frac{\sqrt{3}}{6}$

Using supplementary angles

For $-1<x<0, \theta=\cos ^{-1} x \in\left(\frac{\pi}{2}, \pi\right)$

But, $\sin (\pi-\theta)=\sin \theta={\sin \left(\cos ^{-1} x\right)}$

$\pi-\theta \in(0, \frac{\pi} { 2}) \in$ Range set of sin$^{-1}$

$\sin (\pi-\theta)=\frac{1} {2} .$

$\Rightarrow \pi-\theta=\sin ^{-1} (\frac{1} {2})$

$\pi-\cos ^{-1} x=\sin ^{-1} (\frac{1} {2})=\frac{\pi} {6}$

$\cos^{-1}x=\frac{5\pi} {6}$

$x=\cos \frac{5\pi} {6}=-\cos \frac{\pi} {6}=-\frac{\sqrt{3}} {2}$

$f$ : Trigonometric functions $(\sin , \cos , \tan , \cot$, sec, cosec)

$f: A \longrightarrow B$

$f^{-1}: B \longrightarrow C⊂A$

If $f(\theta)=x, \theta \in C$ then $\theta=f^{-1}(x)$

If $f(\theta)=x, \theta \notin C$ then $\theta=$ ? $\theta \neq f^{-1}(x)$

$\sin : \mathbb{R} \longrightarrow[-1,1]$

$\sin ^{-1}:[-1,1] \longrightarrow[\frac{-\pi} {2}, \frac{\pi} {2}]$

$\sin (\theta)=x, x \in[-1,1]$

Find $\theta, \text { if } \theta \in\left[m \pi-\frac{\pi}{2}, {m \pi +\frac{\pi}{2}}\right]$ , where m is an integer.

$\sin (\theta)=x,$

$m=0, \theta \in [\frac{-\pi} {2}, \frac{\pi} {2}]$

$\theta=\sin ^{-1} x$

$\sin \theta=x, \quad \theta \in\left[m \pi-\frac{\pi}{2}, m \pi-\frac{\pi}{2}\right]$

$\theta-m \pi \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ $\leftarrow$ Range set of sin$^{-1}$

$\sin (\theta-m \pi)=\sin \theta \cos m \pi$

$=(-1)^m$

$=(-1)^m \sin \theta$

$=(-1)^m x$

m even:

$\sin (\theta-m \pi)=x$

$\theta-m \pi=\operatorname{sin}^{-1} x$

$\theta=m \pi+\sin ^{-1} x$

m odd ;

$\sin (\theta-m \pi)=(-1)^m x =-x$

$\theta-m \pi=\sin ^{-1}(-x) =-\sin ^{-1} x$

$\theta=m \pi-\sin ^{-1} x{. }$

In general,

$\theta \in\left[m \pi-\frac{\pi}{2}, m \pi+ \frac{\pi}{2}\right]$ s.t. $\sin \theta=x,|x| \leqslant 1$

$\theta = m \pi + \sin ^{-1} x ,$ m is even

$\theta = m \pi - \sin ^{-1} x ,$ m is odd

Show :

$\tan \left(\tan ^{-1} x+\tan ^{-1} y\right)=\frac{\tan \left(\tan ^{-1} x\right)+\tan \left(\tan ^{-1} y\right)}{1-\tan \left(\tan ^{-1} x\right) \tan \left(\tan ^{-1} y\right)}=\frac{x+y}{1-x y}$

Solution :

Let,

$A=\tan ^{-1} x, ~ ~ B=\tan ^{-1} y$

$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$

$\tan \left(\tan ^{-1} x+\tan ^{-1} y\right)= \frac{x+y}{1-x y}$

$\tan ^{-1} x+\tan ^{-1} y = ?$

In general,

$\tan ^{-1} x+\tan ^{-1} y \neq \tan ^{-1}\left(\frac{x+y}{1-x y}\right)$

$\because\quad$ Range set for tan$^{-1}$ is $(\frac{-\pi}{2},\frac{\pi}{2})$

We know,

$\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}, x \in \mathbb{R}$

$\tan ^{-1}\left(\frac{1}{x}\right)=\cot ^{-1} x, x>0$

Add the above formulas we get,

$\tan ^{-1} x+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{2}, x>0$

$\Rightarrow -\tan ^{-1} x-\tan ^{-1}\left(\frac{1}{x}\right)=\frac{-\pi}{2},x>0$

$\tan ^{-1}(-x)+\tan ^{-1}(\frac{1}{-x})=\frac{-\pi}{2},-x<0$

Put,

$y=-x$

$\tan ^{-1} y+\tan ^{-1} (\frac{1}{y})=\frac{-\pi}{2}, y<0$

Condition-1:

For xy < 1

Case (i)

$x>0, y>0, x y<1$

$\tan ^{-1} x \in\left(0, \frac{\pi}{2}\right) , \tan ^{-1} y \in\left(0, \frac{\pi}{2}\right)$

$0 < y < \frac{1}{x}$

$0<\tan ^{-1} y<\tan ^{-1} (\frac{1}{x})…(a)$

We know, $0<\tan^{-1}x$ (See graph) and add $\tan ^{-1} x$ to equation (a)

$\Rightarrow 0<\tan ^{-1} x<\tan ^{-1} x+\tan ^{-1} y <\tan ^{-1} x+\tan ^{-1} (\frac{1}{x})=\frac{\pi}{2}$

Condition-1: For xy < 1

Case (ii)

$x<0, y<0, x y<1$

$\tan ^{-1} x \in\left(\frac{-\pi}{2}, 0\right) , \tan ^{-1} y \in\left(\frac{-\pi}{2}, 0\right)$

$y>\frac{1}{x}$

$0>\tan ^{-1} y>\tan ^{-1} (\frac{1}{x})…(b)$

We know, $0>\tan^{-1}x$ (See graph) and add $\tan ^{-1} x$ to equation (b)

$\Rightarrow 0>\tan ^{-1} x>\tan ^{-1} x+\tan ^{-1} y>\tan ^{-1} x +\tan ^{-1} \frac{1}{x}$

$(\because\quad \tan ^{-1} x +\tan ^{-1} \frac{1}{x}=\frac{-\pi}{2})$

$\Rightarrow 0>\tan ^{-1} x+\tan ^{-1} y>\frac{-\pi}{2}$

Case (iii)

For xy $\leq$ 0

$x y\leqslant 0$

$x \leqslant 0, y \geqslant 0$

$\tan ^{-1} x \in[\frac{-\pi}{2},0], \tan ^{-1} y \in\left[0, \frac{\pi}{2}\right]$

$\Rightarrow\tan ^{-1} x+\tan ^{-1} y \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

$\quad x \geqslant 0, y \leqslant 0$

$\tan ^{-1} x \in\left[0, \frac{\pi}{2}\right], \tan ^{-1} y \in\left[\frac{-\pi}{2}, 0\right]$

$\tan ^{-1} x+\tan ^{-1} y \in[\frac{-\pi}{2} , \frac{\pi}{2}]$

Condition-2:

$~ ~ x>0, y>0, \quad x y>1$

$\tan ^{-1} x \in (0, \frac{\pi}{2} )$

$\tan ^{-1} y \in(0, \frac{\pi}{2} )$

$y> \frac{1}{x}$

$\Rightarrow \tan ^{-1} y>\tan ^{-1} (\frac{1}{x})$

$\tan ^{-1} x+\tan ^{-1} y>\underbrace{\tan ^{-1} x+\tan ^{-1} \frac{1}{x}}_{=\frac{\pi}{2}}$

$\Rightarrow \tan ^{-1} x+\tan ^{-1} y> \frac{\pi}{2}$

Condition-2:

$\theta=\tan ^{-1} x+\tan ^{-1} y>\frac{\pi}{2}$

$\Rightarrow\theta<\pi ~ ~ ~ \because\tan^{-1} x < \frac{\pi}{2}$ & $\tan^{-1}y<\frac{\pi}{2}$

$\Rightarrow\left.\theta \in (\frac{\pi}{2},\pi\right)\theta ∉$ Range set of $f^{-1}$

But, $\left.\theta-\pi \in (\frac{-\pi}{2}, 0\right)$ ⊂ Range set of $\tan^{-1}$

$\Rightarrow\tan (\theta-\pi)=\tan \theta=\frac{x+y}{1-x y}$

$\theta-\pi = \tan^{-1}\frac{x+y}{1-x y}$

$\rightarrow$ For, $|x|≤ 1$

$2 \tan^{-1} x = \tan^{-1}[\frac{2x}{1-x^2}]$

$\rightarrow$ For, $x > 1$

$2 \tan^{-1} x = \pi + \tan^{-1}[\frac{2x}{1-x^2}]$

$\rightarrow$ For, $x<-1$

$2 \tan^{-1} x = - \pi + \tan^{-1}[\frac{2x}{1-x^2}]$

$\because \tan^{-1}$ is odd function

$\Rightarrow\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1} x+\tan ^{-1}(-y)$

To make identity something like, $\sin ^{-1} x=\tan ^{-1}(?) , |x| \leqslant 1$

Let,$~ \theta=\sin ^{-1} x,\theta \in[-\frac{\pi}{2}, \frac{\pi}{2}]$

$\tan \theta= \frac{\sin \theta }{\cos \theta}= \frac{x}{\sqrt{1-x^2} \quad}$

$\because\cos ^2 \theta =1-\sin ^2 \theta =1-x^2 \Rightarrow \cos \theta= \sqrt{1-x^2}~$because $\cos \theta>0$

$\tan \left(\sin ^{-1} x\right)=\frac{x}{\sqrt{1-x^2}}$

$\Rightarrow\theta=\tan^{-1} (\frac{x}{\sqrt{1-x^2}}) \quad \because \theta = \sin^{-1}x$

$\Rightarrow\sin^{-1}x = \tan^{-1}\frac{x}{\sqrt{1-x^2}}$

In, $~ ~\tan ^{-1}(x)=\sin ^{-1}(?), x \in \mathbb{R}$ we have to find out(?)

Let$~ \theta$, $\theta=\tan ^{-1}(x) \in( \frac{-\pi}{2} , \frac{\pi}{2})$

$\sin \theta=\frac{\tan \theta}{\operatorname{sec} \theta}=\frac{\tan \theta}{\sqrt{1+\tan ^2 \theta}} \because \sec \theta>0$

$= \frac{x}{\sqrt{1+x^2}}$

$\because \ x = \tan\theta 1+ \tan^{2} \theta= \sec^{2} \theta$

$\Rightarrow\sin \theta=\frac{x}{\sqrt{1+x^2}} \quad = \sin ({\tan ^{-1} x}) = \frac{x}{\sqrt{1+x^2}}$

$\Rightarrow\tan ^{-1} x = \sin^{-1} \frac{x} {\sqrt{1+x^2}}$

Problem-2

Find

$\sin ^{-1}\left(\frac{1}{3}\right)+\sin^{-1}\left[\frac{2}{3}\left(1-\frac{1}{\sqrt{8}}\right)\right]=?$

Solution

$\sin ^{-1}(x)=\tan ^{-1} \frac{x}{\sqrt{1-x^2}}$

$\sin ^{-1}\left(\frac{1}{3}\right)=\tan ^{-1}\left(\frac{\frac{1}{3}}{\sqrt{\frac{1-1}{9}}}\right)=\tan ^{-1}\left(\frac{1}{\sqrt{8}}\right)$

$\sin ^{-1}\left(\frac{2}{3}\left(1-\frac{1}{\sqrt{8}}\right)\right)=\tan ^{-1}\left[\frac{\frac{2}{3}(\frac{1-1}{ \sqrt{8}})}{\left.\sqrt{1-\frac{4}{9}(1 \frac{-1}{\sqrt{8}}}^2\right)}\right]$

$=\tan ^{-1} \frac{2(\frac{1-1}{\sqrt{8}})}{\left.\sqrt{9-4\left(1+\frac{1}{8}-\frac{2}{\sqrt{8}}\right.}\right)}$

$\tan ^{-1} \frac{2(1-\frac{1}{\sqrt{8}})}{\sqrt{5-\frac{1}{2}+\frac{8}{\sqrt{8}}}}=\tan ^{-1} \frac{2(1-\frac{1}{\sqrt{8}})}{\sqrt{\frac{9}{2}+\sqrt{8}}}$

$\tan ^{-1}\frac{2(\sqrt{8}-1)}{\sqrt{36+8 \sqrt{8}}}$

$\sin ^{-1} \frac{1}{3}+\sin ^{-1}\left[\frac{2}{3}\left(1-\frac{1}{\sqrt{8}}\right)\right]$

$=\tan ^{-1}\left(\frac{1}{\sqrt{8}}\right)+\tan ^{-1}\left(\frac{(\sqrt{8}-1)}{\sqrt{9+2 \sqrt{8}}}\right)$

$\tan ^{-1}x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}, x y<1$

$=\tan ^{-1}\left(\frac{1}{\sqrt{8}}+\frac{(\sqrt{8}-1)}{\sqrt{9+2 \sqrt{8}}}\right)=\tan ^{-1}(1)$

$\left(1 -\frac{(\sqrt{8-1)}}{\sqrt{72+16 \sqrt{8}}}\right)=\frac{\pi}{4}$