Q.Prove that: $\sin[(n+1)A]\cdot\sin[(n+2)A]+\cos[(n+1)A]\cdot\cos[(n+2)A]=\cos A$

When two angles A and B are combined by addition or subtraction, the trigonometric ratio of the combined (compound) angle can be written in terms of the separate sine, cosine and tangent of A and B. The four building-block identities are $\cos(A-B)=\cos A\cos B+\sin A\sin B$, $\cos(A+B)=\cos A\cos B-\sin A\sin B$, $\sin(A+B)=\sin A\cos B+\cos A\sin B$ and $\sin(A-B)=\sin A\cos B-\cos A\sin B$, proved geometrically using the distance between two points on a unit circle. Dividing the sine formulas by the cosine formulas gives the tangent versions $\tan(A\pm B)=\dfrac{\tan A\pm\tan B}{1\mp\tan A\tan B}$, and taking reciprocals gives the cotangent versions. A very common use is to write an 'unfriendly' angle like $15°$, $75°$ or $105°$ as a sum or difference of the standard angles $30°,45°,60°,90°$ so its exact value can be found. The same identities let us prove identities that mix several angles, and, with $B=\dfrac{\pi}{2}$, they generate the allied-angle results. …