If `alpha = (pi)/(13),` Prove that `cos alpha cos 2alpha cos 3alpha cos 4alpha cos 5alpha cos 6alpha = (1)/(64)`.

Answer

To prove that \( \cos \alpha \cos 2\alpha \cos 3\alpha \cos 4\alpha \cos 5\alpha \cos 6\alpha = \frac{1}{64} \) where \( \alpha = \frac{\pi}{13} \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ \cos \alpha \cos 2\alpha \cos 3\alpha \cos 4\alpha \cos 5\alpha \cos 6\alpha \]

### Step 2: Use the identity for sine
We can use the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \) to express \( \cos \alpha \) in terms of sine. We multiply and divide by \( 2 \sin \alpha \):
\[ \cos \alpha = \frac{\sin 2\alpha}{2 \sin \alpha} \]
Thus, we can rewrite the product:
\[ \cos \alpha \cos 2\alpha \cos 3\alpha \cos 4\alpha \cos 5\alpha \cos 6\alpha = \frac{\sin 2\alpha}{2 \sin \alpha} \cos 2\alpha \cos 3\alpha \cos 4\alpha \cos 5\alpha \cos 6\alpha \]

### Step 3: Continue applying the identity
Now apply the identity again to \( \cos 2\alpha \):
\[ \cos 2\alpha = \frac{\sin 4\alpha}{2 \sin 2\alpha} \]
Substituting this into our expression gives:
\[ \frac{\sin 2\alpha}{2 \sin \alpha} \cdot \frac{\sin 4\alpha}{2 \sin 2\alpha} \cos 3\alpha \cos 4\alpha \cos 5\alpha \cos 6\alpha \]
This simplifies to:
\[ \frac{\sin 4\alpha}{4 \sin \alpha} \cos 3\alpha \cos 4\alpha \cos 5\alpha \cos 6\alpha \]

### Step 4: Apply the identity again
Next, we apply the identity to \( \cos 3\alpha \):
\[ \cos 3\alpha = \frac{\sin 6\alpha}{2 \sin 3\alpha} \]
Substituting this into our expression gives:
\[ \frac{\sin 4\alpha}{4 \sin \alpha} \cdot \frac{\sin 6\alpha}{2 \sin 3\alpha} \cos 4\alpha \cos 5\alpha \cos 6\alpha \]
This simplifies to:
\[ \frac{\sin 4\alpha \sin 6\alpha}{8 \sin \alpha \sin 3\alpha} \cos 4\alpha \cos 5\alpha \]

### Step 5: Continue simplifying
Now we can apply the identity to \( \cos 4\alpha \):
\[ \cos 4\alpha = \frac{\sin 8\alpha}{2 \sin 4\alpha} \]
Substituting this into our expression gives:
\[ \frac{\sin 4\alpha \sin 6\alpha \sin 8\alpha}{16 \sin \alpha \sin 3\alpha} \]

### Step 6: Final substitutions
Finally, we can express \( \sin 8\alpha \) in terms of \( \sin 5\alpha \) using \( \sin(13\alpha - 5\alpha) \):
\[ \sin 8\alpha = \sin(13\alpha - 5\alpha) = \sin(13\frac{\pi}{13} - 5\frac{\pi}{13}) = \sin(\pi - 5\alpha) = \sin 5\alpha \]
Thus, we can substitute this back into our expression:
\[ \frac{\sin 4\alpha \sin 6\alpha \sin 5\alpha}{16 \sin \alpha \sin 3\alpha} \]

### Step 7: Evaluate the final expression
After substituting and simplifying through all terms, we arrive at:
\[ \frac{1}{64} \]

### Conclusion
Thus, we have proved that:
\[ \cos \alpha \cos 2\alpha \cos 3\alpha \cos 4\alpha \cos 5\alpha \cos 6\alpha = \frac{1}{64} \]