How do I prove $\cos(4A)= \cos^4A-6\cos^2A\sin^2A+\sin^4A$?

Question

I had seen a solution elsewhere to this problem but get stuck on understanding the steps- please help!

Answer 1

$\cos(2\cdot 2A)= \cos^2 2A-\sin^2 2A=(\cos^2 A- \sin^2 A)^2-(2\sin A \cos A)^2=\cos^4 A -2\sin^2 A \cos^2 A + \sin^4 A-4 \sin^2 A \cos^2 A$

Answer 2

\begin{align}\cos(4A)&=\cos(2A+2A) \\ &= \cos^2(2A)-\sin^2(2A) \\ &=\cos(2A)\cos(2A)-(2\sin(2A)\cos(2A))^2 \\ &=(\cos^2(2A)-\sin^2(2A))(\cos^2(2A)-\sin^2(2A)) - 4\cos^2(2A)\sin^2(2A) \\ &=(\cos^4(2A)-2\cos^2(2A)\sin^2(2A)+\sin^4(2A))- 4\cos^2(2A)\sin^2(2A) \\ &=\cos^4(2A)-6\cos^2(2A)\sin^2(2A)+\sin^4(2A)\end{align}

Answer 3

$$\cos(4A) + i \sin(4A) = e^{4Ai} = \left(e^{Ai}\right)^4 = (\cos(A) + i\sin(A))^4$$

Expand the binomial and equate the real parts.

I realize this is probably beyond the scope of your curriculum, but I thought you might want to see the easy way to do these trig identities.

Answer 4

Another method: Take second derivatives of both sides to show that $y''=-16y$ on each side. Then show that $\operatorname{left}(0)=\operatorname{right}(0)$ and $\operatorname{left}'(0)=\operatorname{right'}(0)$. This confirms that both sides are the unique solution to the differential equation $y''=-16y$ having some specific initial conditions.