Question:
Find the principal value of:
\[ \sin^{-1}\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right) \]
Solution:
Step 1: Use identity
Recall:
\[ \sin\left(\frac{5\pi}{12}\right) = \frac{\sqrt{6} + \sqrt{2}}{4} \]
Now simplify:
\[ \frac{\sqrt{3}+1}{2\sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{4} \]
Step 2: Substitute
\[ \sin^{-1}\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) \]
Step 3: Check principal range
\[ \frac{5\pi}{12} \notin \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \]
Use identity:
\[ \sin^{-1}(x) = \pi – \theta \quad \text{if } \theta \in \left[\frac{\pi}{2}, \pi\right] \]
Here:
\[ \sin^{-1}\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) = \pi – \frac{5\pi}{12} = \frac{7\pi}{12} \]
But principal value must lie in \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
So correct value is:
\[ \frac{5\pi}{12} \]
—Final Answer:
\[ \boxed{\frac{5\pi}{12}} \]