JEE Preparation From Class 11 - Complete Guide for Early Success

Multiple Angles - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Syllabus

Quick Facts

Multiple angles in terms of arctan and arccos is considered one the most difficult concept.
24 Questions around this concept.

Solve by difficulty

EASY

Considering only the principal values of inverse trigonometric functions, the number of positive real values of x satisfying $\tan ^{-1}(x)+\tan ^{-1}(2 x)=\frac{\pi}{4}$ is :

more than 2

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If $|\mathrm{x}| \leq 1$, then $2 \tan ^2 x+\sin ^{-1} \frac{2 x}{1+x^2}$ is equal to

4 tan–1 x

$\frac{\pi}{2}$

View Solution

Concepts Covered - 4

Multiple angles in terms of arcsin

Multiple angles in terms of arcsin
1. $2 \sin ^{-1} \mathrm{x}=\left\{\begin{array}{cc}\sin ^{-1}\left(2 x \sqrt{1-x^2}\right), & \frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}} \\ \pi-\sin ^{-1}\left(2 x \sqrt{1-x^2}\right), & x>\frac{1}{\sqrt{2}} \\ -\pi-\sin ^{-1}\left(2 x \sqrt{1-x^2}\right), & x<-\frac{1}{\sqrt{2}}\end{array}\right.$
2. $3 \sin ^{-1} \mathrm{x}=\left\{\begin{array}{cc}\sin ^{-1}\left(3 x-4 x^3\right), & -\frac{1}{2} \leq x \leq \frac{1}{2} \\ \pi-\sin ^{-1}\left(3 x-4 x^3\right), & x>\frac{1}{2} \\ -\pi-\sin ^{-1}\left(3 x-4 x^3\right) & x:-\frac{1}{2}\end{array}\right.$

Multiple angles in terms of arccos

Multiple angles in terms of arccos
1. $2 \cos ^{-1} \mathrm{x}=\left\{\begin{array}{cc}\cos ^{-1}\left(2 x^2-1\right), & \text { if } 0 \leq x \leq 1 \\ 2 \pi-\cos ^{-1}\left(2 x^2-1\right), & \text { if }-1 \leq x \leq 0\end{array}\right.$
2. $3 \cos ^{-1} \mathrm{x}=\left\{\begin{array}{cc}\cos ^{-1}\left(4 x^3-3 x\right), & \text { if } \frac{1}{2} \leq x \leq 1 \\ 2 \pi-\cos ^{-1}\left(4 x^3-3 x\right), & \text { if }-\frac{1}{2} \leq x \leq \frac{1}{2} \\ 2 \pi+\cos ^{-1}\left(4 x^3-3 x\right), & \text { if }-1 \leq x \leq-\frac{1}{2}\end{array}\right.$

Multiple angles in terms of arctan and arcsin

$
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

$
2 \tan ^{-1} \mathrm{x}=\left\{\begin{array}{cl}
\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.
$