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[" iv) "sin[(n+1)A]*sin[(n+2)A]+cos[(n+1...

[" iv) "sin[(n+1)A]*sin[(n+2)A]+cos[(n+1)A]],[cos[(n+2)A]=cos A]

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Prove the following: sin[(n+1)A].sin[(n+2)A]+cos[(n+1)A].cos[(n+2)A]=cosA

Prove that: sin(n+1)A sin(n+2)A+cos(n+1)A cos(n+2)A=cos A

Prove that: sin(n+1)As in(n+2)A+cos(+1)A cos(n+2)A=cos A

sin (n +1) x sin (n +2) x + cos (n +1) x cos (n +2) x = cos x

sin (n +1) x sin (n +2) x + cos (n +1) x cos (n +2) x = cos x

sin (n +1) x sin (n +2) x + cos (n +1) x cos (n +2) x = cos x

sin (n +1) x sin (n +2) x + cos (n +1) x cos (n +2) x = cos x

sin (n + 1) A sin (n-1) A + cos (n + 1) A cos (n-1) A = cos2A