Question
Evaluate:
\[ \sin\left(\frac{1}{4}\sin^{-1}\left(\frac{\sqrt{63}}{8}\right)\right) \]
Solution
Let
\[ \theta = \sin^{-1}\left(\frac{\sqrt{63}}{8}\right) \Rightarrow \sin\theta = \frac{\sqrt{63}}{8} \]
Then,
\[ \cos\theta = \sqrt{1 – \frac{63}{64}} = \frac{1}{8} \]
So,
\[ \theta = \cos^{-1}\left(\frac{1}{8}\right) \]
We need:
\[ \sin\left(\frac{\theta}{4}\right) \]
Use half-angle twice:
\[ \cos\frac{\theta}{2} = \sqrt{\frac{1+\cos\theta}{2}} = \sqrt{\frac{1+\frac{1}{8}}{2}} = \sqrt{\frac{9}{16}} = \frac{3}{4} \]
\[ \sin\frac{\theta}{2} = \sqrt{\frac{1-\cos\theta}{2}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4} \]
Now again:
\[ \sin\frac{\theta}{4} = \sqrt{\frac{1 – \cos\frac{\theta}{2}}{2}} = \sqrt{\frac{1 – \frac{3}{4}}{2}} = \sqrt{\frac{1}{8}} = \frac{1}{2\sqrt{2}} \]
Final Answer:
\[ \boxed{\frac{1}{2\sqrt{2}}} \]
Key Concept
Use triangle method and repeated half-angle identities.