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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.

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

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  • $\begingroup$ Thanks so much I got it now $\endgroup$ – Rhia Jan 29 '19 at 20:57
  • $\begingroup$ You are welcome! $\endgroup$ – maxbp Jan 29 '19 at 21:04
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$$f(x)=\sin(30^{\circ}+x)+\sin(30^{\circ}-x)=2\sin30^{\circ}\cos{x}=\cos{x}.$$ Thus, $$f(4x)+4f(2x)=\cos4x+4\cos2x=8\cos^4x-8\cos^2x+1+8\cos^2x-4=8\cos^4x-3.$$

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