Question

The trigonometric equation ${\sin }^{-1}x=2{\sin }^{-1}a$, has a solution for-

• A. $\left|a\right|\le \frac{1}{\sqrt{2}}$
• B. $\frac{1}{2}<\left|a\right|<\frac{1}{\sqrt{2}}$
• C. $allrealvaluesofa$
• D. $\left|a\right|<\frac{1}{2}$

Answer 1

The correct option is A $\left|a\right|\le \frac{1}{\sqrt{2}}$

The trigonometric equation ${\text{sin}}^{-1}x=2{\text{sin}}^{-1}a$ will have a solution if $2{\text{sin}}^{-1}a\in [-\frac{\pi }{2},\frac{\pi }{2}]$ $(\because {\text{sin}}^{-1}x\in [-\frac{\pi }{2},\frac{\pi }{2}\left]\right)$

$⟹{\text{sin}}^{-1}a\in [-\frac{\pi }{4},\frac{\pi }{4}]$

$⟹a\in [-\sin \frac{\pi }{4},\sin \frac{\pi }{4}]$

$⟹a\in [-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]$

$⟹|a|\le \frac{1}{\sqrt{2}}$