To expand $\cos^4 \theta$ using De Moivre's theorem, we express powers of $\cos \theta$ in terms of complex exponentials. First, let's recall De Moivre's theorem:

$\left( \cos \theta + i \sin \theta \right)^n = \cos (n\theta) + i \sin (n\theta).$

Now, introduce $z = e^{i\theta}$, where:

$z = \cos \theta + i \sin \theta \quad \text{and} \quad z^{-1} = \cos \theta - i \sin \theta.$

From this, we know:

$\cos \theta = \frac{z + z^{-1}}{2}.$

Step 1: Express $\cos^4 \theta$ in terms of $z$

Now let's express $\cos^4 \theta$ using this identity:

$\cos^4 \theta = \left( \frac{z + z^{-1}}{2} \right)^4.$

Step 2: Expand the expression

Expand the binomial:

$\cos^4 \theta = \frac{1}{16} \left( z + z^{-1} \right)^4.$

Using the binomial expansion:

$\left( z + z^{-1} \right)^4 = z^4 + 4z^3z^{-1} + 6z^2z^{-2} + 4z z^{-3} + z^{-4}.$

Simplify the powers of $z$:

$\left( z + z^{-1} \right)^4 = z^4 + 4z^2 + 6 + 4z^{-2} + z^{-4}.$

Step 3: Combine terms

Now we substitute this back into the expression for $\cos^4 \theta$:

$\cos^4 \theta = \frac{1}{16} \left( z^4 + 4z^2 + 6 + 4z^{-2} + z^{-4} \right).$

Final Answer:

So, the expansion of $\cos^4 \theta$ in terms of powers of $z$ is:

$\cos^4 \theta = \frac{1}{16} \left( z^4 + 4z^2 + 6 + 4z^{-2} + z^{-4} \right).$

Would you like further details or have any questions about this?

Here are 5 related questions:

1. How can De Moivre's theorem be applied to express $\sin^4 \theta$?
2. Can you derive the expression for $\cos^2 \theta$ using De Moivre's theorem?
3. How does Euler's formula relate to De Moivre's theorem?
4. What happens when expanding $\cos^6 \theta$ in terms of powers of $z$?
5. How can you use De Moivre's theorem to compute $\cos(4\theta)$?

Tip: De Moivre's theorem is especially useful when dealing with powers of trigonometric functions.