Question

Prove the following Sin [(n+1) A] . Sin [(n+2 )A] +cos [(n+1 )A] . Cos [(n+2)A] = cos A

Answer 1

Answer:

Step-by-step explanation:

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%28alpha-beta%29=cos%28alpha%29cos%28beta%29%2Bsin%28alpha%29sin%28beta%29 with alpha=%28n%2B2%29A and beta=%28n%2B1%29A

So the left side becomes:

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

cos[nA + 2A - nA - A]

cos(A)

Hope this helps you

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