Question:
The value of $\sin^{-1}\left(\sin \dfrac{5\pi}{9} \cos \dfrac{\pi}{9} + \sin \dfrac{\pi}{9} \cos \dfrac{5\pi}{9}\right)$ is equal to:
The value of $\sin^{-1}\left(\sin \dfrac{5\pi}{9} \cos \dfrac{\pi}{9} + \sin \dfrac{\pi}{9} \cos \dfrac{5\pi}{9}\right)$ is equal to:
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Always convert expressions using identities before applying inverse functions.
Updated On: Apr 24, 2026
- $\frac{2\pi}{3}$
- $\frac{\pi}{2}$
- $\frac{\pi}{6}$
- $\frac{\pi}{3}$
- $\frac{\pi}{9}$
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The Correct Option is D
Solution and Explanation
Concept:
• $\sin A \cos B + \cos A \sin B = \sin(A+B)$
• Range of $\sin^{-1}x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
Step 1: Apply identity
\[ \sin \frac{5\pi}{9} \cos \frac{\pi}{9} + \sin \frac{\pi}{9} \cos \frac{5\pi}{9} = \sin\left(\frac{5\pi}{9} + \frac{\pi}{9}\right) \] \[ = \sin\left(\frac{6\pi}{9}\right) = \sin\left(\frac{2\pi}{3}\right) \]
Step 2: Evaluate value
\[ \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \]
Step 3: Apply inverse sine
\[ \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} \] Final Conclusion:
\[ = \frac{\pi}{3} \]
• $\sin A \cos B + \cos A \sin B = \sin(A+B)$
• Range of $\sin^{-1}x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
Step 1: Apply identity
\[ \sin \frac{5\pi}{9} \cos \frac{\pi}{9} + \sin \frac{\pi}{9} \cos \frac{5\pi}{9} = \sin\left(\frac{5\pi}{9} + \frac{\pi}{9}\right) \] \[ = \sin\left(\frac{6\pi}{9}\right) = \sin\left(\frac{2\pi}{3}\right) \]
Step 2: Evaluate value
\[ \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \]
Step 3: Apply inverse sine
\[ \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} \] Final Conclusion:
\[ = \frac{\pi}{3} \]
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