Question

Use De Moivre's theorem to express cos 5θ and sin 5θ in terms of sin θ and cos θ.

Answer 1

So, by de Moivre's we know:
\[
(\cos \theta+i \sin\theta)^n=\cos(n\theta)+i\sin(n\theta)
\]Therefore, we know:
\[
\Re\left(\left(\cos\theta+i\sin\theta\right)^5\right)=\cos(5\theta)\\
\Im\left(\left(\cos\theta+i\sin\theta\right)^5\right)=\sin(5\theta)
\]Expand the previous identities, and find the real part to give a closed-form expression for \(\cos\theta\), it follows similarly (for the imaginary part) for \(\sin\theta\).

Answer 2

Would my answer then be (cos θ + i sin θ)^5 = cos 5θ + i sin 5θ?

Answer 3

Although the original answer I gave *is* a correct answer. It should not be the final statement, it was more of a hint. Let's solve for \(\cos(\theta)\) and leave the other for you. Using binomial theorem:
\[
\left(\cos\theta+i\sin(\theta)\right)^5=\\
\cos^5\theta+C_1i(\sin\theta)(\cos^4\theta)+C_2(i^2\sin^2\theta)(\cos^3\theta)+\cdots+C_5(i^5\sin^5\theta)
\]We can ignore all of the odd exponents (excluding the first), since we are looking for the real part, and, thus, we get:
\[
\cos^5\theta-10(\sin^2\theta)(\cos^3\theta)+5(\sin^4\theta)(\cos\theta)=\cos(5\theta)
\]