Represent `cos 6 theta` in terms of `cos theta`.

Explanation

To represent \( \cos 6\theta \) in terms of \( \cos \theta \), we can use the binomial theorem and some trigonometric identities. Here’s a step-by-step solution:

### Step 1: Use Euler's Formula
We start with the expression for \( \cos 6\theta \) using Euler's formula:
\[ \cos 6\theta = \text{Re}(e^{i6\theta}) = \text{Re}((e^{i\theta})^6) = \text{Re}((\cos \theta + i \sin \theta)^6) \]

### Step 2: Expand Using Binomial Theorem
Using the binomial theorem, we expand \( (\cos \theta + i \sin \theta)^6 \):
\[ (\cos \theta + i \sin \theta)^6 = \sum_{k=0}^{6} \binom{6}{k} (\cos \theta)^{6-k} (i \sin \theta)^k \]
This gives us:
\[ = \sum_{k=0}^{6} \binom{6}{k} (\cos \theta)^{6-k} i^k (\sin \theta)^k \]

### Step 3: Separate Real and Imaginary Parts
The real part corresponds to the terms where \( k \) is even:
\[ \cos 6\theta = \binom{6}{0} \cos^6 \theta + \binom{6}{2} \cos^4 \theta (-\sin^2 \theta) + \binom{6}{4} \cos^2 \theta (-\sin^4 \theta) + \binom{6}{6} (-\sin^6 \theta) \]
Calculating the binomial coefficients:
\[ = \cos^6 \theta - 15 \cos^4 \theta \sin^2 \theta + 15 \cos^2 \theta \sin^4 \theta - \sin^6 \theta \]

### Step 4: Substitute \( \sin^2 \theta \)
Using the identity \( \sin^2 \theta = 1 - \cos^2 \theta \):
\[ = \cos^6 \theta - 15 \cos^4 \theta (1 - \cos^2 \theta) + 15 \cos^2 \theta (1 - \cos^2 \theta)^2 - (1 - \cos^2 \theta)^3 \]

### Step 5: Simplify the Expression
Now we simplify the expression:
1. Expand \( \sin^2 \theta \):
- \( \sin^2 \theta = 1 - \cos^2 \theta \)
- \( \sin^4 \theta = (1 - \cos^2 \theta)^2 = 1 - 2\cos^2 \theta + \cos^4 \theta \)
- \( \sin^6 \theta = (1 - \cos^2 \theta)^3 = 1 - 3\cos^2 \theta + 3\cos^4 \theta - \cos^6 \theta \)

2. Substitute back into the equation and combine like terms:
\[ \cos 6\theta = \cos^6 \theta - 15 \cos^4 \theta (1 - \cos^2 \theta) + 15 \cos^2 \theta (1 - 2\cos^2 \theta + \cos^4 \theta) - (1 - 3\cos^2 \theta + 3\cos^4 \theta - \cos^6 \theta) \]

### Step 6: Collecting Terms
After collecting and simplifying all terms, we arrive at:
\[ \cos 6\theta = 32 \cos^6 \theta - 48 \cos^4 \theta + 18 \cos^2 \theta - 1 \]

### Final Result
Thus, the representation of \( \cos 6\theta \) in terms of \( \cos \theta \) is:
\[ \cos 6\theta = 32 \cos^6 \theta - 48 \cos^4 \theta + 18 \cos^2 \theta - 1 \]