Question:
If \( \cos^{-1}x > \sin^{-1}x \), then \(x\) lies in the interval
If \( \cos^{-1}x > \sin^{-1}x \), then \(x\) lies in the interval
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Convert inverse trig inequalities into standard trig inequalities for easier solving.
Updated On: May 8, 2026
- \( \left(\frac{1}{2},1\right] \)
- \( (0,1] \)
- \( [-1,\frac{1}{\sqrt{2}}) \)
- \( [-1,1] \)
- \( [0,1] \)
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The Correct Option is C
Solution and Explanation
Concept:
\[
\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}
\]
Step 1: Rewrite inequality
\[ \cos^{-1}x > \sin^{-1}x \]
Step 2: Substitute
\[ \cos^{-1}x > \frac{\pi}{2} - \cos^{-1}x \]
Step 3: Solve
\[ 2\cos^{-1}x > \frac{\pi}{2} \] \[ \cos^{-1}x > \frac{\pi}{4} \]
Step 4: Convert to cosine
\[ x < \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}} \]
Step 5: Domain
\[ x \in [-1,1] \]
Step 6: Final Answer
\[ \boxed{\left[-1,\frac{1}{\sqrt{2}}\right)} \]
Step 1: Rewrite inequality
\[ \cos^{-1}x > \sin^{-1}x \]
Step 2: Substitute
\[ \cos^{-1}x > \frac{\pi}{2} - \cos^{-1}x \]
Step 3: Solve
\[ 2\cos^{-1}x > \frac{\pi}{2} \] \[ \cos^{-1}x > \frac{\pi}{4} \]
Step 4: Convert to cosine
\[ x < \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}} \]
Step 5: Domain
\[ x \in [-1,1] \]
Step 6: Final Answer
\[ \boxed{\left[-1,\frac{1}{\sqrt{2}}\right)} \]
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