Question

The solution of ${\sin }^{-1}\left(x\right)-{\sin }^{-1}\left(2x\right)=\pm \frac{\mathrm{\pi }}{3}$ is

• A. $\pm \frac{1}{3}$
• B. $\pm \frac{1}{4}$
• C. $\pm \frac{\sqrt{3}}{2}$
• D. $\pm \frac{1}{2}$

Answer 1

The correct option is D

$\pm \frac{1}{2}$

Explanation for the correct option:

Given: ${\sin }^{-1}\left(x\right)-{\sin }^{-1}\left(2x\right)=\pm \frac{\mathrm{\pi }}{3}$

$\Rightarrow {\sin }^{-1}\left(2x\right)={\sin }^{-1}\left(x\right)-{\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)$

As we know that ${\sin }^{-1}a-{\sin }^{-1}b={\sin }^{-1}\left(a\sqrt{1-b^2}-b\sqrt{1-a^2}\right)$

Put $a=x,b=\frac{\sqrt{3}}{2}$in the formula

$$\begin{gathered} \Rightarrow {\sin }^{-1}\left(2x\right)={\sin }^{-1}\left(x\sqrt{1-{\left(\frac{\sqrt{3}}{2}\right)}^2}-\frac{\sqrt{3}}{2}\sqrt{1-{\left(x\right)}^2}\right) \\ \Rightarrow {\sin }^{-1}\left(2x\right)={\sin }^{-1}\left(x\sqrt{1-\frac{3}{4}}-\frac{\sqrt{3}}{2}\sqrt{1-x^2}\right) \\ \Rightarrow {\sin }^{-1}\left(2x\right)={\sin }^{-1}\left(\frac{x}{2}-\frac{\sqrt{3}}{2}\sqrt{1-x^2}\right) \end{gathered}$$

comparing both sides

$$\begin{gathered} \Rightarrow 2x=\frac{x}{2}-\frac{\sqrt{3}}{2}\sqrt{1-x^2} \\ \Rightarrow \frac{3x}{2}=-\frac{\sqrt{3}}{2}\sqrt{1-x^2} \end{gathered}$$

squaring both sides

$$\begin{gathered} \Rightarrow \frac{9x^2}{4}=\frac{3}{4}\left(1-x^2\right) \\ \Rightarrow 3x^2=1-x^2 \\ \Rightarrow x^2=\frac{1}{4} \\ \Rightarrow x=\pm \frac{1}{2} \end{gathered}$$

Hence, option D is correct.