`tan 4 phi ` in terms of ` cos 8phi,`

Answer

To express \( \tan 4\phi \) in terms of \( \cos 8\phi \), we can follow these steps:

### Step 1: Use the Cosine Double Angle Formula
Recall the cosine double angle formula:
\[ \cos 2a = 1 - \tan^2 a \quad \text{or} \quad \cos 2a = \frac{1 - \tan^2 a}{1 + \tan^2 a} \]
Here, we can set \( a = 4\phi \). Thus, we have:
\[ \cos 8\phi = \cos(2 \cdot 4\phi) = \frac{1 - \tan^2(4\phi)}{1 + \tan^2(4\phi)} \]

### Step 2: Rearrange the Equation
We can rearrange the equation to isolate \( \tan^2(4\phi) \):
\[ \cos 8\phi (1 + \tan^2(4\phi)) = 1 - \tan^2(4\phi) \]
Expanding this gives:
\[ \cos 8\phi + \cos 8\phi \tan^2(4\phi) = 1 - \tan^2(4\phi) \]

### Step 3: Collect Like Terms
Rearranging the equation to collect the \( \tan^2(4\phi) \) terms on one side:
\[ \cos 8\phi \tan^2(4\phi) + \tan^2(4\phi) = 1 - \cos 8\phi \]
Factoring out \( \tan^2(4\phi) \):
\[ \tan^2(4\phi)(\cos 8\phi + 1) = 1 - \cos 8\phi \]

### Step 4: Solve for \( \tan^2(4\phi) \)
Now, we can solve for \( \tan^2(4\phi) \):
\[ \tan^2(4\phi) = \frac{1 - \cos 8\phi}{\cos 8\phi + 1} \]

### Step 5: Take the Square Root
Finally, to find \( \tan(4\phi) \), we take the square root of both sides:
\[ \tan(4\phi) = \pm \sqrt{\frac{1 - \cos 8\phi}{\cos 8\phi + 1}} \]

### Final Answer
Thus, we express \( \tan 4\phi \) in terms of \( \cos 8\phi \) as:
\[ \tan 4\phi = \pm \sqrt{\frac{1 - \cos 8\phi}{\cos 8\phi + 1}} \]

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