Compute $$\cos^4 20^{\circ}+\cos^4 40^{\circ}+\cos^4 60^{\circ}+\cos^4 80^{\circ}$$
Suppose that this is a scenario where calculator isn't allowed. I want to say that this expression has something to do with this equation $\cos^2 20^{\circ}+\cos^2 40^{\circ}+\cos^2 60^{\circ}+\cos^2 80^{\circ}=\frac 74$ but I can't seem to find a method to solve this without relying on brute forcing every term to $\cos20^{\circ}$
We can use the "inverse" rule : just like how we compute $\cos n \theta$ from powers of $\cos \theta$, we may compute powers of $\cos \theta$ using values of $\cos n\theta$, which is what we should desire here, since the sequence $20n$ has some regularity(remainder on division by $90$, values for which we know $\sin \backslash\cos$ like $60,120$ etc.) which we can exploit.
Indeed, we have the following situation : $2\cos^2 \theta = \cos 2 \theta - 1$, so squaring and using the $\cos 2\theta$ formula again gives: $$ \cos^4 \theta = \frac{3 + 4\cos 2 \theta + \cos 4 \theta}{8} $$
Let's use this formula: $$ 8\cos^4 20 = 3 + 4 \cos 40 + \cos 80 \\ 8 \cos^4 40 = 3 + 4 \cos 80 + \cos 160 \\ 8 \cos^4 80 = 3 + 4 \cos 160 + \cos 320 $$
Note that $\cos 40 + \cos 80 + \cos 160 = 0$, and $\cos 80 + \cos 160 + \cos 320 = 0$ ! These can be checked by combining two of the three angles into the sum of cosines type formula.
Hence, adding all the above gives $\cos^4 20 + \cos^4 40 + \cos^4 80 = \frac 9{8}$. Adding $\cos^4 60 = \frac 1{16}$ gives the answer $\frac{19}{16}$.
Moral : the inverse formulas are really helpful in problems involving multiples of numbers with "regularity" as I described above.
$$t_1=\cos20^\circ,t_2=-\cos40^\circ,t_3=-\cos80^\circ$$ are the roots of $$8t^3-6t-1=0$$
$$\implies t_1+t_2+t_3=0,t_1t_2+t_2t_3+t_3t_1=-\dfrac68,t_1t_2t_3=\dfrac18$$ and $8t^4=6t^2+t$
$$\implies8(t_1^4+t_2^4+t_3^4)$$
$$=6(t_1^2+t_2^2+t_3^2)+(t_1+t_2+t_3)$$
$$=6((t_1+t_2+t_3)^2-2(t_1t_2+t_2t_3+t_3t_1))$$
$$=-12\cdot-\dfrac68=?$$
See also : Newton–Girard formulae
