To express $\cos(5x)$ using De Moivre's theorem, we first use the formula for expressing powers of complex numbers in polar form. De Moivre's theorem states that for any real number $x$ and integer $n$:

$(\cos(x) + i\sin(x))^n = \cos(nx) + i\sin(nx)$

We want to express $\cos(5x)$, which corresponds to the real part of $(\cos(x) + i\sin(x))^5$. Let's compute this step by step:

1. Begin with the binomial expansion of $(\cos(x) + i\sin(x))^5$:

$(\cos(x) + i\sin(x))^5 = \sum_{k=0}^{5} \binom{5}{k} \cos^{5-k}(x) (i\sin(x))^k$

2. Simplify each term:
   • When $k$ is even, the powers of $i$ are real ($i^2 = -1$).
   • When $k$ is odd, the powers of $i$ are imaginary ($i^3 = -i$, etc.).

This expansion results in the full expression, but we only need the real part to express $\cos(5x)$. This corresponds to the terms where $k$ is even.

Now we apply the binomial expansion specifically to find the real part:

$\cos(5x) = \text{Re}\left((\cos(x) + i\sin(x))^5\right)$

The detailed expansion of this term yields:

$\cos(5x) = 16\cos^5(x) - 20\cos^3(x) + 5\cos(x)$

Thus, the final expression for $\cos(5x)$ using De Moivre's theorem is:

$\cos(5x) = 16\cos^5(x) - 20\cos^3(x) + 5\cos(x)$

Would you like further details or have any questions?

Related Questions:

1. How can we generalize De Moivre's theorem for any $n$?
2. What is the expression for $\sin(5x)$ using De Moivre's theorem?
3. How do we use De Moivre's theorem to compute powers of complex numbers?
4. Can De Moivre's theorem be applied to find roots of complex numbers?
5. What is the relationship between De Moivre's theorem and Euler's formula?

Tip:

De Moivre's theorem is especially useful when raising trigonometric expressions to a power, as it simplifies complex calculations through binomial expansion.