Math, asked by alluvvksatyanarayana, 8 months ago

Cos(n+1)acod(n-1)a+sin(n-1)a

Answers

Answered by Anonymous

10

Answer:

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]

So the left side becomes:

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

cos[nA + 2A - nA - A] cos(A)

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