Show that `sqrt(2+sqrt(2+sqrt(2+2cos 8theta)))=2cos theta`

Answer

To prove that \( \sqrt{2 + \sqrt{2 + \sqrt{2 + 2 \cos 8\theta}}} = 2 \cos \theta \), we will simplify the left-hand side step by step.

### Step 1: Start with the innermost expression
We begin with the innermost expression, which is \( 2 \cos 8\theta \). We can rewrite this using the double angle formula for cosine:
\[ \cos 8\theta = 2 \cos^2 4\theta - 1 \]
Thus, we have:
\[ 2 \cos 8\theta = 2(2 \cos^2 4\theta - 1) = 4 \cos^2 4\theta - 2 \]

### Step 2: Substitute into the expression
Now substitute \( 2 \cos 8\theta \) back into the expression:
\[ \sqrt{2 + \sqrt{2 + \sqrt{2 + 4 \cos^2 4\theta - 2}}} \]
This simplifies to:
\[ \sqrt{2 + \sqrt{2 + 4 \cos^2 4\theta - 2}} = \sqrt{2 + \sqrt{4 \cos^2 4\theta}} \]

### Step 3: Simplify the square root
The expression \( \sqrt{4 \cos^2 4\theta} \) simplifies to \( 2 \cos 4\theta \):
\[ \sqrt{2 + 2 \cos 4\theta} \]

### Step 4: Use the cosine double angle formula again
Now, we can apply the cosine double angle formula again:
\[ \cos 4\theta = 2 \cos^2 2\theta - 1 \]
Thus,
\[ 2 \cos 4\theta = 4 \cos^2 2\theta - 2 \]
Substituting this back gives us:
\[ \sqrt{2 + 2 \cos 4\theta} = \sqrt{2 + 4 \cos^2 2\theta - 2} = \sqrt{4 \cos^2 2\theta} \]

### Step 5: Simplify further
This simplifies to:
\[ 2 \cos 2\theta \]

### Step 6: Repeat the process
Now we repeat the process with \( 2 \cos 2\theta \):
\[ \sqrt{2 + 2 \cos 2\theta} \]
Using the cosine double angle formula again:
\[ \cos 2\theta = 2 \cos^2 \theta - 1 \]
Thus,
\[ 2 \cos 2\theta = 4 \cos^2 \theta - 2 \]
Substituting this back gives us:
\[ \sqrt{2 + 2 \cos 2\theta} = \sqrt{2 + 4 \cos^2 \theta - 2} = \sqrt{4 \cos^2 \theta} \]

### Step 7: Final simplification
This simplifies to:
\[ 2 \cos \theta \]

### Conclusion
Thus, we have shown that:
\[ \sqrt{2 + \sqrt{2 + \sqrt{2 + 2 \cos 8\theta}}} = 2 \cos \theta \]
Hence proved.