Problem
Prove: \[ \frac{9\pi}{8} – \frac{9}{4}\sin^{-1}\left(\frac{1}{3}\right) = \frac{9}{4}\sin^{-1}\left(\frac{2\sqrt{2}}{3}\right) \]
Solution
Step 1: Factor common term
\[ = \frac{9}{4}\left(\frac{\pi}{2} – \sin^{-1}\left(\frac{1}{3}\right)\right) \]
Step 2: Use identity
\[ \frac{\pi}{2} – \sin^{-1}x = \cos^{-1}x \]
\[ = \frac{9}{4}\cos^{-1}\left(\frac{1}{3}\right) \]
Step 3: Show equivalence
Let: \[ \theta = \cos^{-1}\left(\frac{1}{3}\right) \]
\[ \cos\theta = \frac{1}{3} \Rightarrow \sin\theta = \sqrt{1 – \frac{1}{9}} = \frac{2\sqrt{2}}{3} \]
\[ \theta = \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right) \]
Step 4: Substitute
\[ = \frac{9}{4}\sin^{-1}\left(\frac{2\sqrt{2}}{3}\right) \]
Final Result
\[ \boxed{\frac{9}{4}\sin^{-1}\left(\frac{2\sqrt{2}}{3}\right)} \]
Explanation
Using identity π/2 − sin⁻¹x = cos⁻¹x and converting cosine to sine using triangle relation.