# Prove that $\cos(nx)=2\cos((n-1)x)\cos(x)-\cos((n-2)x)$

I'm trying to prove this with the following identities:

$\cos(a)+\cos(b)=2\cos(\frac{a+b}{2})\cos(\frac{a-b}{2})$

$\cos(a)-\cos(b)=-2\cos(\frac{a+b}{2})\cos(\frac{a-b}{2})$

Whenever I try to reduce the term like $2\cos((n-1)x)\cos(x)$ first, I try to set $x = \frac{a-b}{2}$ and use the first equality, but I don't know how to use the expression $\cos((n-1)\frac{a-b}{2})\cos(\frac{a-b}{2})$.

Applying the identity $\cos(a+b)=\cos a\cos b-\sin a\sin b$ we have:

$$$$\cos(nx)=\cos((n-1)x+x)=\cos((n-1)x)\cos(x)-\sin ((n-1)x)\sin x.\tag{1}$$$$

Applying the identity $\cos(a-b)=\cos a\cos b+\sin a\sin b$ we have:

$$$$\cos((n-2)x)=\cos((n-1)x-x)=\cos((n-1)x)\cos(x)+\sin ((n-1)x)\sin x.\tag{2}$$$$

Now, if we add $(1)$ and $(2)$ we get

$$\cos (nx)+\cos((n-2)x)=2\cos((n-1)x)\cos(x),$$ frome where your equality follows.

try the following simultaneous equations:

a/2 + b/2 = (n-1)x

a/2 - b/2 = x

add/subtract equations to get values for a and b, which you can then use on the LHS of your identity.

This is so straightforward. Let's try using a = nx and b = (n-2)x in the first identity:

$cos(a) + cos(b) = 2cos(\frac{a+b}{2})cos(\frac{a-b}{2})$

$\Rightarrow cos(nx) + cos((n-2)x) = 2cos(\frac{nx+(n-2)x}{2})cos(\frac{nx-(n-2)x}{2})$

$\Rightarrow cos(nx) + cos((n-2)x) = 2cos((n-1)x)cos(x)$

$\Rightarrow cos(nx) = 2cos((n-1)x)cos(x) - cos((n-2)x)$