Question:medium

If $x \in (0, 1)$ and $\sin^{-1} x - \sin^{-1} \frac{1}{4} = \frac{\pi}{3}$, then $x$ equals

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When dealing with inverse trigonometric equations, substituting $\theta = \sin^{-1} y$ and drawing a right triangle can make finding $\cos\theta$ intuitive and less prone to errors.
Updated On: Jun 16, 2026
  • $\frac{1}{8}(1 + 3\sqrt{5})$
  • $\frac{1}{6}(1 + 2\sqrt{5})$
  • $\frac{1}{6}(1 + 4\sqrt{3})$
  • $\frac{1}{5}(2 + 3\sqrt{2})$
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The Correct Option is A

Solution and Explanation

Step 1: Move one inverse sine across.
Start with $\sin^{-1} x - \sin^{-1} \frac{1}{4} = \frac{\pi}{3}$. Rearrange to $\sin^{-1} x = \frac{\pi}{3} + \sin^{-1} \frac{1}{4}$.

Step 2: Take sine of both sides.
Then $x = \sin\left(\frac{\pi}{3} + \sin^{-1} \frac{1}{4}\right)$.

Step 3: Use the sine addition rule.
With $A = \frac{\pi}{3}$ and $B = \sin^{-1} \frac{1}{4}$, we have $x = \sin A \cos B + \cos A \sin B$.

Step 4: Fill in the known values.
Here $\sin A = \frac{\sqrt{3}}{2}$ and $\cos A = \frac{1}{2}$. For angle $B$, $\sin B = \frac{1}{4}$, so $\cos B = \sqrt{1 - \frac{1}{16}} = \frac{\sqrt{15}}{4}$ (positive, since $B$ is a small first-quadrant angle).

Step 5: Plug everything in.
$x = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{15}}{4} + \frac{1}{2} \cdot \frac{1}{4} = \frac{\sqrt{45}}{8} + \frac{1}{8} = \frac{3\sqrt{5}}{8} + \frac{1}{8}$.

Step 6: Combine into a single fraction.
$x = \frac{1 + 3\sqrt{5}}{8} = \frac{1}{8}(1 + 3\sqrt{5})$, which lies in $(0, 1)$ as required.
\[ \boxed{x = \tfrac{1}{8}\left(1 + 3\sqrt{5}\right)} \]
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