Rewriting $\cos 6t-4\cos 4t+7\cos 2t $ as $4\cos^3(2t)-8\cos^2(2t)+4\cos2t+4$

Question

I have an assignment question that has me completely stumped.
$$ h=\cos6t-4\cos4t+7\cos2t $$ where I need to show it can be rewritten as $$ h=4\cos^3(2t)-8\cos^2(2t)+4\cos2t+4 $$ I can work out $\cos6t$ to equal $2\cos^2(3t)-1$ but I have no idea on how to use the trig identities for the other sections of the equation with numbers in front.

Can anyone shed some light on this for me. In basic terms? Thanks!

Answer 1

Hint:

Use the duplication formula $$\cos 2\theta=2\cos^2\theta-1$$ and derive the formula for $$\cos3\theta=\cos(2\theta+\theta)$$ (unless you've already learnt the formula for it).

Answer 2

Answer (readers, feel free to check and critique!)

I found it easier working from the middle first, then doing a substitution for the first and third terms.

For the middle term $$-4 \cos 4t$$ we have

$$-4 \cos 4t \Rightarrow -4 (\cos 2(2t)) \Rightarrow -4(2 \cos^2(2t) - 1) \Rightarrow -8 \cos^2 (2t) + 4.$$

For the terms $\cos 6t$ and $7 \cos 2t$, let $z = 2t.$ That is,

$$\cos 6t + 7 \cos 2t \Rightarrow \cos 3(2t) + 7 \cos 2t \Rightarrow \cos 3z + 7 \cos z.$$

Since $\cos 3z = 4 \cos^3 z - 3 \cos z$, we add this to $7 \cos z$ to get

$$4 \cos^3 z + 4 \cos z.$$

then resubstituting $z = 2t$ we have $$4 \cos^3 2t + 4 \cos 2t.$$

Adding $4 \cos^3 2t + 4 \cos 2t$ to $-8 \cos^2 (2t) + 4$ and rearranging gives us the final answer of

$$4 \cos^3 2t -8 \cos^2 (2t) + 4 \cos 2t + 4$$ $$\mathbf {QED}$$