Answer to Question #282428 in Complex Analysis for iqa

Question

Apply De Moivre’s formula to express cos 4θ and sin 4θ in terms of cos θ and sin θ.

Answer 1

Let us apply De Moivre’s formula to express $\cos 4θ$ and $\sin 4θ$ in terms of $\cos θ$ and $\sin θ.$ It follows from De Moivre’s formula that $(\cos\theta+i\sin\theta)^4=\cos4\theta+i\sin4\theta,$ where $i^2=-1.$

On the other hand, the Binomial Theorem implies that $(\cos\theta+i\sin\theta)^4 \\=\cos^4\theta+4\cos^3\theta\cdot i\sin\theta+6\cos^2\theta\cdot i^2\sin^2\theta +4\cos\theta\cdot i^3\sin^3\theta+i^4\sin^4\theta \\=\cos^4\theta+4i\cos^3\theta\cdot \sin\theta-6\cos^2\theta\cdot \sin^2\theta -4i\cos\theta\cdot \sin^3\theta+\sin^4\theta \\=\cos^4\theta-6\cos^2\theta\cdot \sin^2\theta +\sin^4\theta+i(4\cos^3\theta\cdot \sin\theta-4\cos\theta\cdot \sin^3\theta).$

Taking into account that two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal, we conclude that

$\cos4\theta=\cos^4\theta-6\cos^2\theta\cdot \sin^2\theta +\sin^4\theta, \\ \sin4\theta=4\cos^3\theta\cdot \sin\theta-4\cos\theta\cdot \sin^3\theta.$