Inverse trigonometric ratios of Multiple Angles

Prior to understanding the multiple angles of inverse trigonometric function, let's discuss inverse trigonometric function. Inverse trigonometric functions can be defined as the inverses of the basic trigonometric functions - sine, cosine, tangent, cotangent, secant, and cosecant. The multiple-angle formulas for these trigonometric functions can also be represented in the form of their inverse functions.

In this article, we will cover the concept of Multiple angles. This category falls under the broader category of Trigonometry, which is a crucial Chapter in class 12 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. A total of six questions have been asked on this topic including one in 2014, one in 2019, two in 2021, and two in 2023.

What are Multiple angles in terms of inverse trigonometric function?

The multiple angle formula is the representation of the angle which is multiple of a given angle. We can calculate the values of multiple angles by expressing each trigonometric function in its expanded form.

Multiple angles in terms of arcsin

1. Double of Inverse Trigonometric Function Formulas

$2 \sin ^{-1} \mathrm{x}=\left\{\begin{array}{cc}\sin ^{-1}\left(2 x \sqrt{1-x^2}\right), & \frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}} \\ \pi-\sin ^{-1}\left(2 x \sqrt{1-x^2}\right), & x>\frac{1}{\sqrt{2}} \\ -\pi-\sin ^{-1}\left(2 x \sqrt{1-x^2}\right), & x<-\frac{1}{\sqrt{2}}\end{array}\right.$
2. Triple of Inverse Trigonometric Function Formulas

$3 \sin ^{-1} \mathrm{x}=\left\{\begin{array}{cc}\sin ^{-1}\left(3 x-4 x^3\right), & -\frac{1}{2} \leq x \leq \frac{1}{2} \\ \pi-\sin ^{-1}\left(3 x-4 x^3\right), & x>\frac{1}{2} \\ -\pi-\sin ^{-1}\left(3 x-4 x^3\right) & x:-\frac{1}{2}\end{array}\right.$

Multiple angles in terms of arccos

1. Double of Inverse Trigonometric Function Formulas

$2 \cos ^{-1} \mathrm{x}=\left\{\begin{array}{cc}\cos ^{-1}\left(2 x^2-1\right), & \text { if } 0 \leq x \leq 1 \\ 2 \pi-\cos ^{-1}\left(2 x^2-1\right), & \text { if }-1 \leq x \leq 0\end{array}\right.$
2. Triple of Inverse Trigonometric Function Formulas

$3 \cos ^{-1} \mathrm{x}=\left\{\begin{array}{cc}\cos ^{-1}\left(4 x^3-3 x\right), & \text { if } \frac{1}{2} \leq x \leq 1 \\ 2 \pi-\cos ^{-1}\left(4 x^3-3 x\right), & \text { if }-\frac{1}{2} \leq x \leq \frac{1}{2} \\ 2 \pi+\cos ^{-1}\left(4 x^3-3 x\right), & \text { if }-1 \leq x \leq-\frac{1}{1}\end{array}\right.$

Multiple angles in terms of arctan and arcsin

The multiple-angle formula of arctan in terms of arcsin is given by

$2 \tan ^{-1} \mathrm{x}=\left\{\begin{array}{cc}\sin ^{-1}\left(\frac{2 x}{1+x^2}\right), & \text { if }-1 \leq x \leq 1 \\ \pi-\sin ^{-1}\left(\frac{2 x}{1+x^2}\right), & \text { if } x>1 \\ -\pi-\sin ^{-1}\left(\frac{2 x}{1+x^2}\right), & \text { if } x<-1\end{array}\right.$

Multiple angles in terms of arctan and arccos

The multiple angle formula of arctan in terms of arccos is given by

$2 \tan ^{-1} \mathrm{x}=\left\{\begin{array}{cc}\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right), & \text { if } 0 \leq x<\infty \\ -\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right), & \text { if }-\infty<x \leq 0\end{array}\right.$

Summary

The multiple angle formula of the inverse trigonometric function allows manipulation of multiple angles of the trigonometric function in terms of the inverse trigonometric function of the original angle. These formulas are used in various types of problems involving inverse trigonometric functions. Hence, knowledge about these problems provides insight into the relationship between angles and trigonometric functions.

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Solved Examples Based on Multiple Angles in Terms of Inverse Trigonometric Functions

Example 1: $S=\left\{x \in R ; 0<x<1\right.$ and $\left.2 \tan ^{-1}\left(\frac{1-x}{1+x}\right)=\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right\}$ If n(S) denotes the number of elements in S then : [JEE MAINS 2023]

Solution: Put $x=\tan \theta \quad \theta \in\left(0, \frac{\pi}{4}\right)$

$\begin{aligned} & 2 \tan ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right)=\cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right) \\ & 2 \tan ^{-1}\left[\tan \left(\frac{\pi}{4}-\theta\right)\right]=\cos ^{-1}[\cos (2 \theta)] \\ & \Rightarrow 2\left(\frac{\pi}{4}-\theta\right)=2 \theta \Rightarrow \theta=\frac{\pi}{8} \\ & \Rightarrow x=\tan \frac{\pi}{8}=\sqrt{2}-1 \simeq 0.414\end{aligned}$

Hence, the answer is $n(S)=2$ and only one element in S is less then $\frac{1}{2}$

Example 2: If the sum of all the solutions of $\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)+\cot ^{-1}\left(\frac{1-x^2}{2 x}\right)=\frac{\pi}{3},-1<x<1, x \neq 0{ }_{\text {is }} \alpha-\frac{4}{\sqrt{3}}$, then $\alpha$ is equal to
[JEE MAINS 2023]

Solution$x \in(-1,1) \quad \tan ^{-1}\left(\frac{2 \mathrm{x}}{1-\mathrm{x}^2}\right)=2 \tan ^{-1} \mathrm{x}$
$x \in(0,1) \quad \cot ^{-1}\left(\frac{1-x^2}{2 x}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)=2 \tan ^{-1} x$
$x \in(-1,0) \quad \cot ^{-1}\left(\frac{1-x^2}{2 x}\right)=\mathrm{IT}+\tan ^{-1}\left(\frac{2 x}{1-\mathrm{x}^2}\right)=\mathrm{IT}+2 \tan ^{-1} \mathrm{x}$$x \in(0,1) \quad 2 \tan ^{-1} x+2 \tan ^{-1} x=\pi / 3$

$\tan ^{-1} x=\frac{\pi}{12}$

$\mathrm{x}=2-\sqrt{3}$$\begin{gathered}x \in(-1,0) \quad 2 \tan ^{-1} x+I T+2 \tan ^{-1} x=\pi / 3 \\ 4 \tan ^{-1} x=\frac{-2 \pi}{3} \\ \tan ^{-1} \mathrm{x}=-\pi / 6 \\ \mathrm{x}=-1 / \sqrt{3} \\ (2-\sqrt{3})+\left(-\frac{1}{\sqrt{3}}\right)=a-\frac{4}{\sqrt{3}} \\ 2-4 / \sqrt{3}=\alpha-4 / \sqrt{3} \\ \alpha=2\end{gathered}$
Hence, the answer is 2.

Example 3: $\operatorname{cosec}\left[2 \cot ^{-1}(5)+\cos ^{-1}\left(\frac{4}{5}\right)\right]$ is equal to: [JEE MAINS 2021]

Solution
$
\begin{aligned}
& \operatorname{cosec}\left[2 \tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{3}{4}\right)\right] \\
& \operatorname{cosec}\left[\tan ^{-1}\left(\frac{5}{12}\right)+\tan ^{-1}\left(\frac{3}{4}\right)\right] \\
& \operatorname{cosec}\left[\tan ^{-1}\left(\frac{\frac{5}{12}+\frac{3}{4}}{1-\frac{5}{12} \times \frac{3}{4}}\right)\right] \\
& =\operatorname{cosec}\left[\tan ^{-1}\left(\frac{56}{33}\right)\right]=\frac{65}{56}
\end{aligned}
$
$
\text { Hence, the answer is } \frac{65}{56}
$

Example 4: The sum of possible values of x for $\tan ^{-1}(x+1)+\cot ^{-1}\left(\frac{1}{x-1}\right)=\tan ^{-1}\left(\frac{8}{31}\right)$ is : [JEE MAINS 2021]

Solution
$
\tan ^{-1}(x+1)+\cot ^{-1}\left(\frac{1}{x-1}\right)=\tan ^{-1} \frac{8}{31}
$
Taking tangent both sides :-

$
\begin{aligned}
& \frac{(x+1)+(x-1)}{1-\left(x^2-1\right)}=\frac{8}{31} \\
& \Rightarrow \frac{2 x}{2-x^2}=\frac{8}{31} \\
& \Rightarrow 4 x^2+31 x-8=0 \\
& \Rightarrow x=-8, \frac{1}{4}
\end{aligned}
$

$\begin{aligned} & \text { But, if } x=\frac{1}{4} \\ & \tan ^{-1}(x+1) \in\left(0, \frac{\pi}{2}\right) \\ & \& \cot ^{-1}\left(\frac{1}{x-1}\right) \in\left(\frac{\pi}{2}, \pi\right)\end{aligned}$
$
\Rightarrow L H S>\frac{\pi}{2} \quad \& \quad R H S<\frac{\pi}{2}
$

(Not possible)
Hence, $x=-8$

Hence, the answer is $-\frac{32}{4}$

Example 5: If $2 y=\left(\cot ^{-1}\left(\frac{\sqrt{3} \cos x+\sin x}{\cos x-\sqrt{3} \sin x}\right)\right)^2, x \in\left(0, \frac{\pi}{2}\right)$ $\square$ then $\overline{d x}$ is equal to [JEE MAINS 2019]

Solution: Important Results of Inverse Trigonometric Functions $\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$ when $x \in R$

$2 y=\left(\cot ^{-1}\left(\frac{\sqrt{3} \cos x+\sin x}{\cos x-\sqrt{3} \sin x}\right)\right)^2 \quad x \in\left(0, \frac{\pi}{2}\right)$

$2 y=\left(\cot ^{-1}\left(\frac{\frac{\sqrt{3} \cos x}{\cos x}+\frac{\sin x}{\cos x}}{\frac{\cos x}{\cos x}-\frac{\sqrt{3} \sin x}{\cos x}}\right)\right)^2$

$=\left(\cot ^{-1}\left(\frac{\sqrt{3}+\tan x}{1-\sqrt{3} \tan x}\right)\right)^2$

$=\left(\cot ^{-1}\left(\tan \left(\frac{\pi}{3}+x\right)\right)\right)^2$

$=\left(\frac{\pi}{2}-\tan ^{-1}\left(\tan \left(\frac{\pi}{3}+x\right)\right)\right)^2$

$=\left(\frac{\pi}{2}-\left(\frac{\pi}{3}+x\right)\right)^2$

$=\left(\frac{\pi}{2}-\frac{\pi}{3}-x\right)^2=\left(\frac{\pi}{6}-x\right)^2$

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

$\frac{d y}{d x}=\frac{1}{2} \times 2\left(x-\frac{\pi}{6}\right)$

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

Hence, the answer is $x-\frac{\pi}{6}$