Chapter 2: Inverse Trigonometric Functions (Relations and Functions)

Let ${\sin }^{-1}\left(-\frac{1}{2}\right)=y$, so $\sin y=-\frac{1}{2}=\sin \left(-\frac{\pi }{6}\right)$. Since $-\frac{\pi }{6}\in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$, the principal value is $-\frac{\pi }{6}$.

$-\frac{\pi }{6}$

$\cos y=\frac{\sqrt{3}}{2}=\cos \frac{\pi }{6}$ and $\frac{\pi }{6}\in [0,\pi ]$.

$\frac{\pi }{6}$

$\text{cosec}y=2=\text{cosec}\frac{\pi }{6}$ and $\frac{\pi }{6}\in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]-\{0\}$.

$\frac{\pi }{6}$

$\tan y=-\sqrt{3}=\tan \left(-\frac{\pi }{3}\right)$ and $-\frac{\pi }{3}\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$.

$-\frac{\pi }{3}$

$\cos y=-\frac{1}{2}=\cos \left(\pi -\frac{\pi }{3}\right)=\cos \frac{2\pi }{3}$ and $\frac{2\pi }{3}\in [0,\pi ]$.

$\frac{2\pi }{3}$

$\tan y=-1=\tan \left(-\frac{\pi }{4}\right)$ and $-\frac{\pi }{4}\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$.

$-\frac{\pi }{4}$

$\sec y=\frac{2}{\sqrt{3}}$, i.e. $\cos y=\frac{\sqrt{3}}{2}=\cos \frac{\pi }{6}$, and $\frac{\pi }{6}\in [0,\pi ]-\left\{\frac{\pi }{2}\right\}$.

$\frac{\pi }{6}$

$\cot y=\sqrt{3}=\cot \frac{\pi }{6}$ and $\frac{\pi }{6}\in (0,\pi )$.

$\frac{\pi }{6}$

$\cos y=-\frac{1}{\sqrt{2}}=\cos \left(\pi -\frac{\pi }{4}\right)=\cos \frac{3\pi }{4}$ and $\frac{3\pi }{4}\in [0,\pi ]$.

$\frac{3\pi }{4}$

$\text{cosec}y=-\sqrt{2}$, i.e. $\sin y=-\frac{1}{\sqrt{2}}=\sin \left(-\frac{\pi }{4}\right)$, and $-\frac{\pi }{4}$ is in the principal branch.

$-\frac{\pi }{4}$

$${\tan }^{-1}1+{\cos }^{-1}\left(-\frac{1}{2}\right)+{\sin }^{-1}\left(-\frac{1}{2}\right)=\frac{\pi }{4}+\frac{2\pi }{3}-\frac{\pi }{6}=\frac{3\pi +8\pi -2\pi }{12}=\frac{9\pi }{12}=\frac{3\pi }{4}$$

$\frac{3\pi }{4}$

$${\cos }^{-1}\frac{1}{2}+2{\sin }^{-1}\frac{1}{2}=\frac{\pi }{3}+2\cdot \frac{\pi }{6}=\frac{2\pi }{3}$$

$\frac{2\pi }{3}$

The principal value branch of ${\sin }^{-1}$ is $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$.

(B)

${\tan }^{-1}\sqrt{3}=\frac{\pi }{3}$. ${\sec }^{-1}(-2)$: $\cos y=-\frac{1}{2}$, so $y=\frac{2\pi }{3}$.

$$\frac{\pi }{3}-\frac{2\pi }{3}=-\frac{\pi }{3}$$

(B)

Let $x=\sin \theta$, so $\theta ={\sin }^{-1}x\in \left[-\frac{\pi }{6},\frac{\pi }{6}\right]$ for $x\in \left[-\frac{1}{2},\frac{1}{2}\right]$. Then

$$3x-4x^3=3\sin \theta -4{\sin }^3\theta =\sin 3\theta$$

Since $3\theta \in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$, ${\sin }^{-1}(\sin 3\theta )=3\theta =3{\sin }^{-1}x$.

With $x=\sin \theta$, $3x-4x^3=\sin 3\theta$ and $3\theta$ lies in the principal range, so the identity holds.

Let $x=\cos \theta$, $\theta ={\cos }^{-1}x\in \left[0,\frac{\pi }{3}\right]$. Then

$$4x^3-3x=4{\cos }^3\theta -3\cos \theta =\cos 3\theta$$

Since $3\theta \in [0,\pi ]$, ${\cos }^{-1}(\cos 3\theta )=3\theta =3{\cos }^{-1}x$.

With $x=\cos \theta$, $4x^3-3x=\cos 3\theta$ and $3\theta \in [0,\pi ]$.

Put $x=\tan \theta$, $\theta \in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$, $\theta \ne 0$. Then $\sqrt{1+x^2}=\sec \theta$ and

$$\frac{\sec \theta -1}{\tan \theta }=\frac{1-\cos \theta }{\sin \theta }=\frac{2{\sin }^2\frac{\theta }{2}}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}=\tan \frac{\theta }{2}$$

So the expression $={\tan }^{-1}\left(\tan \frac{\theta }{2}\right)=\frac{\theta }{2}=\frac{1}{2}{\tan }^{-1}x$.

$\frac{1}{2}{\tan }^{-1}x$

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

For $0<x<\pi$, $\tan \frac{x}{2}>0$, so the square root is $\tan \frac{x}{2}$ and the expression $={\tan }^{-1}\left(\tan \frac{x}{2}\right)=\frac{x}{2}$.

$\frac{x}{2}$

Divide numerator and denominator by $\cos x$:

$$\frac{1-\tan x}{1+\tan x}=\frac{\tan \frac{\pi }{4}-\tan x}{1+\tan \frac{\pi }{4}\tan x}=\tan \left(\frac{\pi }{4}-x\right)$$

Since $-\frac{\pi }{2}<\frac{\pi }{4}-x<\frac{\pi }{2}$, the expression $=\frac{\pi }{4}-x$.

$\frac{\pi }{4}-x$

Put $x=a\sin \theta$, $\theta \in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$. Then $\sqrt{a^2-x^2}=a\cos \theta$ and

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

So the expression $=\theta ={\sin }^{-1}\frac{x}{a}$.

${\sin }^{-1}\frac{x}{a}$

Put $x=a\tan \theta$, with $-\frac{\pi }{6}<\theta <\frac{\pi }{6}$:

$$\frac{3a^3\tan \theta -a^3{\tan }^3\theta }{a^3-3a^3{\tan }^2\theta }=\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }=\tan 3\theta$$

Since $3\theta \in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$, the expression $=3\theta =3{\tan }^{-1}\frac{x}{a}$.

$3{\tan }^{-1}\frac{x}{a}$

${\sin }^{-1}\frac{1}{2}=\frac{\pi }{6}$, so $2{\sin }^{-1}\frac{1}{2}=\frac{\pi }{3}$ and $2\cos \frac{\pi }{3}=2\cdot \frac{1}{2}=1$.

${\tan }^{-1}1=\frac{\pi }{4}$.

$\frac{\pi }{4}$

For $|x|<1$, ${\sin }^{-1}\frac{2x}{1+x^2}=2{\tan }^{-1}x$; for $y>0$, ${\cos }^{-1}\frac{1-y^2}{1+y^2}=2{\tan }^{-1}y$.

$$\tan \frac{1}{2}\left[2{\tan }^{-1}x+2{\tan }^{-1}y\right]=\tan \left({\tan }^{-1}x+{\tan }^{-1}y\right)=\tan \left({\tan }^{-1}\frac{x+y}{1-xy}\right)=\frac{x+y}{1-xy}$$

$\frac{x+y}{1-xy}$

$\frac{2\pi }{3}$ is not in $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$. Write $\sin \frac{2\pi }{3}=\sin \left(\pi -\frac{2\pi }{3}\right)=\sin \frac{\pi }{3}$, and $\frac{\pi }{3}$ is in the range.

$\frac{\pi }{3}$

$\tan \frac{3\pi }{4}=\tan \left(\pi -\frac{\pi }{4}\right)=-\tan \frac{\pi }{4}=\tan \left(-\frac{\pi }{4}\right)$, and $-\frac{\pi }{4}\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$.

$-\frac{\pi }{4}$

${\sin }^{-1}\frac{3}{5}={\tan }^{-1}\frac{3}{4}$ (a 3-4-5 triangle) and ${\cot }^{-1}\frac{3}{2}={\tan }^{-1}\frac{2}{3}$.

$$\tan \left({\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{2}{3}\right)=\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\cdot \frac{2}{3}}=\frac{\frac{17}{12}}{\frac{1}{2}}=\frac{17}{6}$$

$\frac{17}{6}$

$\cos \frac{7\pi }{6}=\cos \left(2\pi -\frac{7\pi }{6}\right)=\cos \frac{5\pi }{6}$, and $\frac{5\pi }{6}\in [0,\pi ]$.

(B)

${\sin }^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi }{6}$, so the expression $=\sin \left(\frac{\pi }{3}+\frac{\pi }{6}\right)=\sin \frac{\pi }{2}=1$.

(D)

${\tan }^{-1}\sqrt{3}=\frac{\pi }{3}$; ${\cot }^{-1}(-\sqrt{3})=\pi -{\cot }^{-1}\sqrt{3}=\pi -\frac{\pi }{6}=\frac{5\pi }{6}$.

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

(B)

$\cos \frac{13\pi }{6}=\cos \left(2\pi +\frac{\pi }{6}\right)=\cos \frac{\pi }{6}$, and $\frac{\pi }{6}\in [0,\pi ]$.

$\frac{\pi }{6}$

$\tan \frac{7\pi }{6}=\tan \left(\pi +\frac{\pi }{6}\right)=\tan \frac{\pi }{6}$, and $\frac{\pi }{6}\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$.

$\frac{\pi }{6}$

Let $\theta ={\sin }^{-1}\frac{3}{5}$, so $\tan \theta =\frac{3}{4}$ and $0<\theta <\frac{\pi }{4}$ (since $\frac{3}{4}<1$).

$$\tan 2\theta =\frac{2\cdot \frac{3}{4}}{1-\frac{9}{16}}=\frac{\frac{3}{2}}{\frac{7}{16}}=\frac{24}{7}$$

Since $0<2\theta <\frac{\pi }{2}$, $2\theta ={\tan }^{-1}\frac{24}{7}$.

$\tan (2\theta )=\frac{24}{7}$ with $2\theta$ in the principal range.

${\sin }^{-1}\frac{8}{17}={\tan }^{-1}\frac{8}{15}$ (8-15-17 triangle) and ${\sin }^{-1}\frac{3}{5}={\tan }^{-1}\frac{3}{4}$.

$${\tan }^{-1}\frac{8}{15}+{\tan }^{-1}\frac{3}{4}={\tan }^{-1}\frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15}\cdot \frac{3}{4}}={\tan }^{-1}\frac{\frac{77}{60}}{\frac{36}{60}}={\tan }^{-1}\frac{77}{36}$$

(valid since $\frac{8}{15}\cdot \frac{3}{4}<1$).

Convert both to ${\tan }^{-1}$ and add: ${\tan }^{-1}\frac{77}{36}$.

Let $A={\cos }^{-1}\frac{4}{5}$ ($\sin A=\frac{3}{5}$) and $B={\cos }^{-1}\frac{12}{13}$ ($\sin B=\frac{5}{13}$); both are acute.

$$\cos (A+B)=\cos A\cos B-\sin A\sin B=\frac{4}{5}\cdot \frac{12}{13}-\frac{3}{5}\cdot \frac{5}{13}=\frac{48-15}{65}=\frac{33}{65}$$

Since $0<A+B<\pi$, $A+B={\cos }^{-1}\frac{33}{65}$.

$\cos (A+B)=\frac{33}{65}$.

Let $A={\cos }^{-1}\frac{12}{13}$ ($\sin A=\frac{5}{13}$) and $B={\sin }^{-1}\frac{3}{5}$ ($\cos B=\frac{4}{5}$).

$$\sin (A+B)=\sin A\cos B+\cos A\sin B=\frac{5}{13}\cdot \frac{4}{5}+\frac{12}{13}\cdot \frac{3}{5}=\frac{20+36}{65}=\frac{56}{65}$$

$A+B$ is acute (both angles are less than $\frac{\pi }{4}$), so $A+B={\sin }^{-1}\frac{56}{65}$.

$\sin (A+B)=\frac{56}{65}$.

${\sin }^{-1}\frac{5}{13}={\tan }^{-1}\frac{5}{12}$ and ${\cos }^{-1}\frac{3}{5}={\tan }^{-1}\frac{4}{3}$.

$${\tan }^{-1}\frac{5}{12}+{\tan }^{-1}\frac{4}{3}={\tan }^{-1}\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}\cdot \frac{4}{3}}={\tan }^{-1}\frac{\frac{21}{12}}{\frac{4}{9}}={\tan }^{-1}\frac{63}{16}$$

(valid since $\frac{5}{12}\cdot \frac{4}{3}=\frac{5}{9}<1$).

Sum of ${\tan }^{-1}\frac{5}{12}$ and ${\tan }^{-1}\frac{4}{3}$ is ${\tan }^{-1}\frac{63}{16}$.

Let $x={\tan }^2\theta$, $\theta \in \left[0,\frac{\pi }{4}\right]$, so $\sqrt{x}=\tan \theta$.

$$\frac{1-x}{1+x}=\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }=\cos 2\theta$$

Since $2\theta \in \left[0,\frac{\pi }{2}\right]$: $\frac{1}{2}{\cos }^{-1}(\cos 2\theta )=\theta ={\tan }^{-1}\sqrt{x}$.

With $x={\tan }^2\theta$, the right side is $\frac{1}{2}(2\theta )=\theta ={\tan }^{-1}\sqrt{x}$.

$1\pm \sin x={\left(\cos \frac{x}{2}\pm \sin \frac{x}{2}\right)}^2$. For $x\in \left(0,\frac{\pi }{4}\right)$, $\cos \frac{x}{2}>\sin \frac{x}{2}>0$, so

$\sqrt{1+\sin x}=\cos \frac{x}{2}+\sin \frac{x}{2}$ and $\sqrt{1-\sin x}=\cos \frac{x}{2}-\sin \frac{x}{2}$.

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

(since $\frac{x}{2}\in (0,\pi )$).

The fraction simplifies to $\cot \frac{x}{2}$, so the expression is $\frac{x}{2}$.

Put $x=\cos 2\theta$; for $-\frac{1}{\sqrt{2}}\le x\le 1$, $\theta \in \left[0,\frac{3\pi }{8}\right]$. Then $\sqrt{1+x}=\sqrt{2}\cos \theta$, $\sqrt{1-x}=\sqrt{2}\sin \theta$:

$$\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }=\frac{1-\tan \theta }{1+\tan \theta }=\tan \left(\frac{\pi }{4}-\theta \right)$$

Since $\frac{\pi }{4}-\theta \in \left[-\frac{\pi }{8},\frac{\pi }{4}\right]$, the expression $=\frac{\pi }{4}-\theta =\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}x$.

With $x=\cos 2\theta$ the fraction is $\tan \left(\frac{\pi }{4}-\theta \right)$, giving $\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}x$.

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

Equating with ${\tan }^{-1}\frac{2}{\sin x}$:

$$\frac{2\cos x}{{\sin }^{2}x}=\frac{2}{\sin x}\Rightarrow \cos x=\sin x\Rightarrow \tan x=1\Rightarrow x=\frac{\pi }{4}$$

$x=\frac{\pi }{4}$

$${\tan }^{-1}\frac{1-x}{1+x}={\tan }^{-1}1-{\tan }^{-1}x=\frac{\pi }{4}-{\tan }^{-1}x$$

So $\frac{\pi }{4}-{\tan }^{-1}x=\frac{1}{2}{\tan }^{-1}x$, giving $\frac{3}{2}{\tan }^{-1}x=\frac{\pi }{4}$, ${\tan }^{-1}x=\frac{\pi }{6}$, and $x=\frac{1}{\sqrt{3}}$.

$x=\frac{1}{\sqrt{3}}$

Let ${\tan }^{-1}x=\theta$, $\tan \theta =x$: a right triangle with opposite $x$, adjacent 1, hypotenuse $\sqrt{1+x^2}$. So $\sin \theta =\frac{x}{\sqrt{1+x^2}}$.

(D)

Let ${\sin }^{-1}x=\theta$. Then ${\sin }^{-1}(1-x)=\frac{\pi }{2}+2\theta$, so

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

$2x^2-x=0$, so $x=0$ or $x=\frac{1}{2}$.

Check $x=\frac{1}{2}$: ${\sin }^{-1}\frac{1}{2}-2{\sin }^{-1}\frac{1}{2}=-\frac{\pi }{6}\ne \frac{\pi }{2}$, rejected. Check $x=0$: ${\sin }^{-1}1-0=\frac{\pi }{2}$ ✔.

(C) x = 0