Question

The value of $\cos (n+1)\alpha \cos (n-1)\alpha +\sin (n+1)\alpha \sin (n-1)\alpha$ is

• A. $\cos 2n\alpha$
• B. $\sin 2n\alpha$
• C. $\cos 2\alpha$
• D. $\sin 2\alpha$

Answer 1

The correct option is C $\cos 2\alpha$

$\cos (n+1)\alpha \cos (n-1)\alpha +\sin (n+1)\alpha \sin (n-1)\alpha$
Assuming
$(n+1)\alpha =A,(n-1)\alpha =B$

$\cos (n+1)\alpha \cos (n-1)\alpha +\sin (n+1)\alpha \sin (n-1)\alpha =\cos A\cos B+\sin A\sin B=\cos (A-B)=\cos \left(\right(n+1)\alpha -(n-1\left)\alpha \right)=\cos 2\alpha$