Question

Sin(n+1)x.sin(n+2)x + cos(n-1)x.cos(n+2)x=cosx
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plz do this sum
class 11 maths

Answer 1

HELLO DEAR,



cos(n+1)xcos(n+2)x + sin(n+1)xsin(n+2)x

it is from the identity

cos(a-b) = cosa * cosb + sina * sinb

here a= (n+1)x b= (n+2)x

so,

cos((n+1)x-(n+2)x)

cos(nx+x-nx-2x)

cos(-x)

cosx { since cos(-x) = cosx }


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Answer:

Answer:

Step-by-step explanation:

Given,

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

Before starting let,

=> a = (n+1)x

=> b = (n+2)x

We know that,

=> cos(a-b) = sin(a).sin(b) + cos(b).cos(a)

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

=> cos((n+1)x - (n+2)x)

=> cos(-x)

=> cos(x)

Hence Proved