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



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:

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

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

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

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:

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

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

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.