Question 551032
<pre>
sin[(n+1)A]sin[(n+2)A] + cos[(n+1)A]cos[(n+2)A] = cosA

Use the commutative principles of multiplication and addition
to rearrange the left side as

cos[(n+2)A]cos[(n+1)A] + sin[(n+2)A]sin[(n+1)A]

The left side is the right side of the identity 
                {{{cos(alpha-beta)=cos(alpha)cos(beta)+sin(alpha)sin(beta)}}} with {{{alpha=(n+2)A}}} and {{{beta=(n+1)A}}}

So the left side becomes:

cos[(n+2)A - (n+1)A]

cos[nA + 2A - nA - A]

cos(A)

Edwin</pre>