First slide

Properties of ITF

Question

If $x \in [- \frac{1}{2}, 1],$ then $\sin^{- 1} (\frac{\sqrt{3}}{2} x - \frac{1}{2} \sqrt{1 - x^{2}})$ , equals

Moderate

A

$\sin^{- 1} \frac{1}{2} - \sin^{- 1} x$
B

$\sin^{- 1} x - \frac{π}{6}$
C

$\sin^{- 1} x + \frac{π}{6}$
D

none of these

Solution

Let $x = \sin θ$ . Then,
$- \frac{1}{2} \leq x \leq 1 \Rightarrow - \frac{1}{2} \leq \sin θ \leq 1 \Rightarrow - \frac{π}{6} \leq θ \leq \frac{π}{2}$
Now,
$\begin{array}{l} \sin^{- 1} (\frac{\sqrt{3}}{2} x - \frac{1}{2} \sqrt{1 - x^{2}}) \\ = \sin^{- 1} (\sin θ \cos \frac{π}{6} - \cos θ \sin \frac{π}{6}) \\ = \sin^{- 1} \{\sin (θ - \frac{π}{6})\} \\ = θ - \frac{π}{6} [∵ - \frac{π}{6} \leq θ \leq \frac{π}{2} \Rightarrow - \frac{π}{3} \leq θ - \frac{π}{6} \leq \frac{π}{3}] \end{array}$
$= \sin^{- 1} x - \frac{π}{6}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App