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Sin((n+1)x)×sin((n+2)x)+cos((n+1)x)×cos((n+2)x)

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Sin((n+1)x)×sin((n+2)x)+cos((n+1)x)×cos((n+2)x)

Posted by Hemant Prajapati 7 years, 8 months ago

CBSE > Class 11 > Mathematics

2 answers

Yash Thakur 7 years, 8 months ago

sin [(n+1)x)] × sin [(n+2)x] + cos [(n+1)x]×cos[(n+2)x]=? SOLUTION------ sin (nx+x) × sin (nx+2x) + cos (nx+x) × cos (nx+2x) = cos (nx+2x) × cos (nx+x) + sin (nx+2x) × sin (nx+x). We know that--- cosA × cosB + sinA × sinB = cos (A-B) So we put value of A is (nx+ 2x) and value of B is (nx+x). cos (A-B)= cos ( nx+2x -nx-x )= cos (2x-x) = cos x ( Ans.)

Pranavam Pranavam 7 years, 8 months ago

Lhs. It is in the form of cos(x+y) Cos ((n-1)x-(n+2)x) Cos (x(n-1-n-2)) Cos(x(-1) Cos(-x) Cos x

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