The value of sin [(n+1) A] sin [(n+2) A+
cos [(n+1)A]. cos[(n+2) A] =
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Step-by-step explanation:
sin [(n+1) A] sin [(n+2) A+
cos [(n+1)A]. cos[(n+2) A] =
Above expression is just like
SinA SinB + CosA CosB = Cos(A-B)
sin [(n+1) A] sin [(n+2) A+
cos [(n+1)A]. cos[(n+2) A] =
Cos [((n+1)-(n+2))A]
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