# Showing that $f(4x) + 4f(2x) =8\cos^4(x)-3$, where $f(x) = \cos x$

It is given that $$f(x)= \sin(x+30^\circ) + \cos(x+60^\circ)$$

A) Show that $$f(x)= \cos(x)$$

B) Hence, show that $$f(4x) + 4f(2x) =8\cos^4(x)-3$$

I managed to prove $$f(x)$$ equals $$\cos(x)$$, but after that I'm stumped.

• Do you know how to expand $\cos 4x$ and $\cos 2x$ in terms of $\cos x$? Even easier if you know Euler's formula. – rogerl Jan 29 '19 at 20:38

You can use the property: $$cos(2x) = cos^2(x)-sin^2(x)$$
Since this will allow you to express $$cos(4x)$$ = $$cos^2(2x)-(1-cos^2(2x)) = 2cos^2(2x)-1$$
By substituting again $$cos(2x)$$ by the aforementioned expression you will be able to express $$4cos(2x)+cos(4x)$$ in terms of $$sin(x)$$ and $$cos(x)$$ raised to some powers. The sinus will cancell out and will obtained the desired result.
$$f(x)=\sin(30^{\circ}+x)+\sin(30^{\circ}-x)=2\sin30^{\circ}\cos{x}=\cos{x}.$$ Thus, $$f(4x)+4f(2x)=\cos4x+4\cos2x=8\cos^4x-8\cos^2x+1+8\cos^2x-4=8\cos^4x-3.$$