Question

prove that sin(n+1)x sin(n+2)x + cos(n+1)x cos(n+2)x =cosx

Answer 1

let (n+1)x be A and (n+2)x be B
then apply the formula cos(a-b) = sin a sin b + cos a cos b
by doing that you'll get cos[nx +x-nx-2x] = cos x
since cos(-x) = cos x
therefore cos x = cos x

Answer 2

I'm going to break this down into two parts and then add them:
The first part:
\[\sin[(n+1)x]\sin[(n+2)x]\]
use the identity:
\[\sin \alpha \sin \beta=\frac{1}{2}[\cos(\alpha-b)-\cos (\alpha+\beta)]\]
in our case:
\[\alpha= (n=1)x, \beta=(n+2)x\]
Thus:
\[\sin[(n+1)x]\sin[(n+2)x]=\]
\[\frac{1}{2}[\cos[(n+1)x-(n+2)x]-\cos[(n+1)x+(n+2)x]]\]
\[=\frac{1}{2}[\cos(-x)-\cos(2nx+3x)]\]
The second part:
\[\cos[(n+1)x]\cos[(n+2)x]\]
use the identity:
\[\cos \alpha \cos \beta=\frac{1}{2}[\cos(\alpha-\beta)+\cos(\alpha+\beta)]\]
in our case:
\[\alpha=(n+1)x, \beta=(n+2)x\]
Thus:
\[\cos[(n+1)x]\cos[(n+2)x]=\]
\[\frac{1}{2}[\cos[(n+1)x-(n+2)x]+\cos[(n+1)x+(n+2)x]]\]
\[=\frac{1}{2}[\cos(-x)+\cos(2nx+3x)]\]
Now add parts one and two:
\[\frac{1}{2}[\cos(-x)-\cos(2nx+3x)]+\frac{1}{2}[\cos(-x)+\cos(2nx+3x)]=\]
\[\frac{1}{2}cos(-x)-\frac{1}{2}\cos(2nx+3x)+\frac{1}{2}cos(-x)+\frac{1}{2}\cos(2nx+3x)=\]
\[\frac{1}{2}\cos(-x)+\frac{1}{2}\cos(-x)=\cos(-x)\]
Since cosine is an even function: cos(-x)=cos(x)