Answer to Question #113437 in Trigonometry for Caylin

Question

Using de Moivres theorem, express

Cos5A and sin4A as polynomials in sin A and cos A.

Answer 1

De Moivre's Theorem:

For any complex number $z=r(\cos \theta +i \sin \theta)$ and any integer $n$,

$z^n=r^n(\cos (n\theta)+i \sin(n\theta))$

We will consider $z=\cos A+i \sin A$

$z^5=\cos 5A+i \sin 5A$

$z^5=(\cos A+i \sin A)^5=\cos^5A+5\cos ^4 A(i \sin A)+10\cos ^3 A(i \sin A)^2+10\cos ^2A(i \sin A)^3+5\cos A(i\sin A)^4+(i \sin A)^5=(\cos ^5A-10\cos ^3 A\sin^2 A+5\cos A\sin^4A)+i(5\cos^4 A\sin A-10\cos^2 A\sin ^3 A+\sin^5A)$

Therefore, $\cos 5A= \cos ^5A-10\cos ^3 A\sin^2 A+5\cos A\sin^4A$

$z^4=\cos 4A+i \sin 4A$

$z^4=(\cos A+i \sin A)^4=\cos ^4A+4\cos ^3A(i\sin A)+6\cos^2 A(i\sin A)^2+4\cos A(i \sin A)^3+(i\sin A)^4=(\cos^4A-6\cos^2 A\sin^2A+\sin^4A)+i(4\cos ^3A\sin A-4\cos A\sin^3A)$

Therefore, $\sin 4A= 4\cos ^3A\sin A-4\cos A\sin^3A$

Answer: $\cos 5A= \cos ^5A-10\cos ^3 A\sin^2 A+5\cos A\sin^4A$

$\sin 4A= 4\cos ^3A\sin A-4\cos A\sin^3A$