Question:
If $x \in (0, 1)$ and $\sin^{-1} x - \sin^{-1} \frac{1}{4} = \frac{\pi}{3}$, then $x$ equals
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: Understanding the Question:
The question asks us to find the value of $x \in (0, 1)$ satisfying a given trigonometric equation involving inverse sine functions.
We will solve this by isolating the term $\sin^{-1} x$ and applying sine to both sides.
Step 2: Key Formula or Approach:
We will use the sine angle addition identity:
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]
And the basic relationship:
\[ \cos(\sin^{-1}\theta) = \sqrt{1 - \theta^2} \text{ for } \theta \in [0, 1] \]
Step 3: Detailed Explanation:
• Let us write down the given equation:
\[ \sin^{-1} x - \sin^{-1} \frac{1}{4} = \frac{\pi}{3} \]
• Rearranging the terms to isolate $\sin^{-1} x$:
\[ \sin^{-1} x = \frac{\pi}{3} + \sin^{-1} \frac{1}{4} \]
• Now, take the sine of both sides of the equation:
\[ x = \sin\left(\frac{\pi}{3} + \sin^{-1} \frac{1}{4}\right) \]
• Let $\theta = \sin^{-1} \frac{1}{4}$. This means $\sin\theta = \frac{1}{4}$.
Since $\theta$ is in the first quadrant, we can find $\cos\theta$:
\[ \cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \left(\frac{1}{4}\right)^2} = \sqrt{1 - \frac{1}{16}} = \frac{\sqrt{15}}{4} \]
• Now substitute this back into the expansion formula for $\sin\left(\frac{\pi}{3} + \theta\right)$:
\[ x = \sin\frac{\pi}{3} \cos\theta + \cos\frac{\pi}{3} \sin\theta \]
• Substituting the values of the trigonometric terms:
\[ x = \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{15}}{4}\right) + \left(\frac{1}{2}\right)\left(\frac{1}{4}\right) \]
• Multiplying and simplifying:
\[ x = \frac{\sqrt{45}}{8} + \frac{1}{8} \]
\[ x = \frac{3\sqrt{5} + 1}{8} = \frac{1}{8}(1 + 3\sqrt{5}) \]
Step 4: Final Answer:
The value of $x$ is $\frac{1}{8}(1 + 3\sqrt{5})$.
The question asks us to find the value of $x \in (0, 1)$ satisfying a given trigonometric equation involving inverse sine functions.
We will solve this by isolating the term $\sin^{-1} x$ and applying sine to both sides.
Step 2: Key Formula or Approach:
We will use the sine angle addition identity:
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]
And the basic relationship:
\[ \cos(\sin^{-1}\theta) = \sqrt{1 - \theta^2} \text{ for } \theta \in [0, 1] \]
Step 3: Detailed Explanation:
• Let us write down the given equation:
\[ \sin^{-1} x - \sin^{-1} \frac{1}{4} = \frac{\pi}{3} \]
• Rearranging the terms to isolate $\sin^{-1} x$:
\[ \sin^{-1} x = \frac{\pi}{3} + \sin^{-1} \frac{1}{4} \]
• Now, take the sine of both sides of the equation:
\[ x = \sin\left(\frac{\pi}{3} + \sin^{-1} \frac{1}{4}\right) \]
• Let $\theta = \sin^{-1} \frac{1}{4}$. This means $\sin\theta = \frac{1}{4}$.
Since $\theta$ is in the first quadrant, we can find $\cos\theta$:
\[ \cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \left(\frac{1}{4}\right)^2} = \sqrt{1 - \frac{1}{16}} = \frac{\sqrt{15}}{4} \]
• Now substitute this back into the expansion formula for $\sin\left(\frac{\pi}{3} + \theta\right)$:
\[ x = \sin\frac{\pi}{3} \cos\theta + \cos\frac{\pi}{3} \sin\theta \]
• Substituting the values of the trigonometric terms:
\[ x = \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{15}}{4}\right) + \left(\frac{1}{2}\right)\left(\frac{1}{4}\right) \]
• Multiplying and simplifying:
\[ x = \frac{\sqrt{45}}{8} + \frac{1}{8} \]
\[ x = \frac{3\sqrt{5} + 1}{8} = \frac{1}{8}(1 + 3\sqrt{5}) \]
Step 4: Final Answer:
The value of $x$ is $\frac{1}{8}(1 + 3\sqrt{5})$.
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