How do I show that $\cos^4x=\frac{1}{8}\cos(4x)+\frac{1}{2}\cos(2x )+\frac{3}{8}$

Question

I know how to prove that

$$\cos^2x=\frac{1}{2}+\frac{1}{2}\cos(2x)$$

by substituting $\cos(2x)$ with $2\cos^2x-1$ according to the double angle identity $$\cos(2x)=2\cos^2x-1$$

However, how do I do that for $\cos^4x$?

Do I do it by writing $\cos^4x$ as $$\cos^2(x)\cdot \cos^2(x)$$ and thus get it by squaring the LHS of $$\cos^2x=\frac{1}{2}+\frac{1}{2}\cos(2x)$$

Im not sure how to proceed. Any ideas?

Answer 1

Since $$\cos^2x=\frac{1}{2}+\frac{1}{2}\cos(2x)$$ we have $$\cos^2(2x)=\frac{1}{2}+\frac{1}{2}\cos(4x)$$ and thus, if we square first equation we get $$\cos^4x=\frac{1}{4}+{1\over 2}\cos(2x)+{1\over 4}\cos ^2(2x)=$$

$$ =\frac{1}{4}+{1\over 2}\cos(2x)+{1\over 4}\Big{(}\frac{1}{2}+\frac{1}{2}\cos(4x)\Big{)}$$ $$ =\frac{3}{8}+{1\over 2}\cos(2x)+\frac{1}{8}\cos(4x)$$

Answer 2

Hint: By squaring we get $$\cos^4(x)=\frac{1}{4}(\cos^2(2x)+2\cos(2x)+1)$$ and then

$$\cos(4x)=2\cos^2(2x)-1$$

Answer 3

As an alternative we can use that

$$\cos x=\Re(e^{ix})$$$$\implies \cos 4x=\Re(e^{i4x})=\Re[(\cos x+i\sin x)^4]=\cos^4 x-6\cos^2x \sin^2x+\sin^4x$$

that is

$$\cos 4x=\cos^4 x-6\cos^2x \sin^2x+\sin^4x$$

$$\cos 4x=\cos^4 x-6\cos^2x (1-\cos^2 x)+(1-\cos^2 x)^2$$

$$\cos 4x=\cos^4 x-6\cos^2x+6\cos^4x+1-2\cos^2x+\cos^4 x$$

$$8\cos^4 x=\cos 4x+8\cos^2x-1$$

$$8\cos^4 x=\cos 4x+8\left(\frac{1}{2}+\frac{1}{2}\cos(2x)\right)-1$$

$$8\cos^4 x=\cos 4x+4\cos(2x)+3$$

$$\cos^4 x=\frac18\cos 4x+\frac12\cos(2x)+\frac38$$