Answer
See proof below.
Work Step by Step
We know that
$\cos(0)=\cos (2 \pi) =\cos (4 \pi) =......=1$
are even values for $k$. That is, $\cos (k \pi)=1$.
We also know that
$\cos(\pi)=\cos (3 \pi) =\cos (5 \pi) =......=-1$
are odd values for $k$. That is, $\cos (k \pi)=-1$.
So, $\cos(k \pi)=(-1)^k$
Apply the sum formula : $\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$
In order to prove the given identity, we simplify the left hand side $\text{LHS}$ as follows:
$\text{LHS}=\cos (\theta +k \pi) =\cos \theta \cos (k \pi) -\sin \theta \sin (k \pi) \\=\cos \theta \cos (k \pi) \\=(-1)^k \cos \theta\\=\text{ RHS}$
Where $k$ is an integer.
