Question

Prove that $\sin (n+1)x\sin (n+2)x+\cos (n+1)x\cos (n+2)x=\cos x$

Answer 1

$L.H.S.=\sin (n+1)x\sin (n+2)x+\cos (n+1)x\cos (n+2)x$

$=\frac{1}{2}\left[2\sin \right(n+1\left)x\sin \right(n+2)x+2\cos (n+1\left)x\cos \right(n+2\left)x\right]$

$=\frac{1}{2}\left[\begin{array}{c}\cos \left\{\right(n+1)x-(n+2\left)x\right\}-cis\left\{\right(n+1)x+(n+2\left)x\right\}\\ +\cos \left\{\right(n+1)x+(n+2\left)x\right\}+\cos \left\{\right(n+1)x-(n+2\left)x\right\}\end{array}\right]$

$\left[\begin{array}{c}\because -2\sin A\sin B=\cos (A+B)-\cos (A-B)\\ 2\cos A\cos B=\cos (A+B)+\cos (A-B)\end{array}\right]$

$=\frac{1}{2}\times 2\cos \left\{\right(n+1)x-(n+2\left)x\right\}$

$=\cos (-x)=\cos x=R.H.S.$