Evaluate `tan(1/4*Sin^-1(sqrt(63)/8))`

Answer

To solve the expression \( \tan\left(\frac{1}{4} \sin^{-1}\left(\frac{\sqrt{63}}{8}\right)\right) \), we will follow these steps:

### Step 1: Set up the equation
Let \( x = \sin^{-1}\left(\frac{\sqrt{63}}{8}\right) \). This implies that:
\[ \sin x = \frac{\sqrt{63}}{8} \]

### Step 2: Find \( \cos x \)
Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \), we can find \( \cos x \):
\[ \cos^2 x = 1 - \sin^2 x = 1 - \left(\frac{\sqrt{63}}{8}\right)^2 = 1 - \frac{63}{64} = \frac{1}{64} \]
Thus,
\[ \cos x = \sqrt{\frac{1}{64}} = \frac{1}{8} \]

### Step 3: Use the half-angle formula
Now we need to find \( \tan\left(\frac{x}{4}\right) \). We can use the half-angle formulas:
\[ \tan\left(\frac{x}{4}\right) = \frac{\sin\left(\frac{x}{4}\right)}{\cos\left(\frac{x}{4}\right)} \]
Using the half-angle identities:
\[ \sin\left(\frac{x}{2}\right) = \sqrt{\frac{1 - \cos x}{2}} \quad \text{and} \quad \cos\left(\frac{x}{2}\right) = \sqrt{\frac{1 + \cos x}{2}} \]

### Step 4: Calculate \( \sin\left(\frac{x}{2}\right) \) and \( \cos\left(\frac{x}{2}\right) \)
Substituting \( \cos x = \frac{1}{8} \):
\[ \sin\left(\frac{x}{2}\right) = \sqrt{\frac{1 - \frac{1}{8}}{2}} = \sqrt{\frac{\frac{7}{8}}{2}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4} \]
\[ \cos\left(\frac{x}{2}\right) = \sqrt{\frac{1 + \frac{1}{8}}{2}} = \sqrt{\frac{\frac{9}{8}}{2}} = \sqrt{\frac{9}{16}} = \frac{3}{4} \]

### Step 5: Calculate \( \tan\left(\frac{x}{2}\right) \)
Now we can find \( \tan\left(\frac{x}{2}\right) \):
\[ \tan\left(\frac{x}{2}\right) = \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} = \frac{\frac{\sqrt{7}}{4}}{\frac{3}{4}} = \frac{\sqrt{7}}{3} \]

### Step 6: Calculate \( \tan\left(\frac{x}{4}\right) \)
Using the half-angle formula again:
\[ \tan\left(\frac{x}{4}\right) = \sqrt{\frac{1 - \cos\left(\frac{x}{2}\right)}{1 + \cos\left(\frac{x}{2}\right)}} \]
Substituting \( \cos\left(\frac{x}{2}\right) = \frac{3}{4} \):
\[ \tan\left(\frac{x}{4}\right) = \sqrt{\frac{1 - \frac{3}{4}}{1 + \frac{3}{4}}} = \sqrt{\frac{\frac{1}{4}}{\frac{7}{4}}} = \sqrt{\frac{1}{7}} = \frac{1}{\sqrt{7}} \]

### Final Answer
Thus, the value of \( \tan\left(\frac{1}{4} \sin^{-1}\left(\frac{\sqrt{63}}{8}\right)\right) \) is:
\[ \frac{1}{\sqrt{7}} \]