View source for 2cos(mt)cos(nt)=cos((m+n)t)+cos((m-n)t)

==Theorem==
The following formula holds for $m,n \in \{0,1,2,\ldots\}$:
$$2\cos(mt)\cos(nt)=\cos((m+n)t)+\cos((m-n)t),$$
where $\cos$ denotes [[cosine]].

==Proof==

==References==
* {{BookReference|An Introduction to Orthogonal Polynomials|1978|T.S. Chihara|next=Orthogonality relation for cosine on (0,pi)}} $(1.1)$

[[Category:Theorem]]
[[Category:Unproven]]