Question

Considering the principal values of the inverse trigonometric functions, $\sin^{-1} \left( \frac{\sqrt{3}}{2} x + \frac{1}{2} \sqrt{1-x^2} \right)$, $-\frac{1}{2}<x<\frac{1}{\sqrt{2}}$, is equal to

• A. $\frac{\pi}{4} + \sin^{-1} x$
• B. $\frac{\pi}{6} + \sin^{-1} x$
• C. $\frac{-5\pi}{6} - \sin^{-1} x$
• D. $\frac{5\pi}{6} - \sin^{-1} x$

Hint:

Use the angle addition formula for inverse trigonometric functions.

Answer 1

The Correct Option is B

1. Let $\sin^{-1} x = \theta$: \[ x = \sin \theta \]
2. Express the given function: \[ \sin^{-1} \left( \frac{\sqrt{3}}{2} x + \frac{1}{2} \sqrt{1-x^2} \right) \] \[ = \sin^{-1} \left( \frac{\sqrt{3}}{2} \sin \theta + \frac{1}{2} \cos \theta \right) \]
3. Use the angle addition formula: \[ = \sin^{-1} \left( \sin \theta \cos \frac{\pi}{6} + \cos \theta \sin \frac{\pi}{6} \right) \] \[ = \sin^{-1} \left( \sin \left( \theta + \frac{\pi}{6} \right) \right) \] \[ = \theta + \frac{\pi}{6} \] \[ = \sin^{-1} x + \frac{\pi}{6} \] Therefore, the correct answer is (2) $\frac{\pi}{6} + \sin^{-1} x$.

Answer 2

We need to simplify the principal value \( \sin^{-1}\!\left( \frac{\sqrt{3}}{2}\,x + \frac{1}{2}\sqrt{1-x^2} \right) \) for \( -\tfrac{1}{2} < x < \tfrac{1}{\sqrt{2}} \).

Concept Used:

Use the identity \(a\sin\theta + b\cos\theta = R\sin(\theta+\phi)\) where \(R=\sqrt{a^2+b^2}\), \( \cos\phi=\frac{a}{R} \), \( \sin\phi=\frac{b}{R}\). Also set \(x=\sin\theta\) so that \(\sqrt{1-x^2}=\cos\theta\) when \(\theta\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]\).

Step-by-Step Solution:

Step 1: Parameterize \(x\) by \(x=\sin\theta\) with \(\theta=\sin^{-1}x\in\left(-\tfrac{\pi}{6},\,\tfrac{\pi}{4}\right)\) 

Step 2: Write as a single sine:

\[ \frac{\sqrt{3}}{2}\sin\theta+\frac{1}{2}\cos\theta = \sin\theta\cdot\cos\frac{\pi}{6}+\cos\theta\cdot\sin\frac{\pi}{6} = \sin\!\left(\theta+\frac{\pi}{6}\right). \]

Step 3: Check principal range for arcsin.

\[ \theta\in\left(-\frac{\pi}{6},\,\frac{\pi}{4}\right)\;\Rightarrow\; \theta+\frac{\pi}{6}\in\left(0,\,\frac{5\pi}{12}\right)\subset\left[-\frac{\pi}{2},\frac{\pi}{2}\right]. \] Thus, \[ \sin^{-1}\!\left(\sin\!\left(\theta+\frac{\pi}{6}\right)\right)=\theta+\frac{\pi}{6}. \]

Final Computation & Result

\[ \sin^{-1}\!\left( \frac{\sqrt{3}}{2}\,x + \frac{1}{2}\sqrt{1-x^2} \right) = \sin^{-1}x + \frac{\pi}{6}. \]

Answer: \( \boxed{\,\sin^{-1}x + \dfrac{\pi}{6}\,} \)