Question

sin⁡7x+6sin⁡5x+17sin⁡3x+12sin⁡xsin⁡6x+5sin⁡4x+12sin⁡2x equals:

• A. (A) sin⁡x
• B. (B) 2sin⁡x
• C. (C) cos⁡x
• D. (D) 2cos⁡x

Answer 1

Answer: (D)

(D) 2cos⁡x

Answer 2

Explanation:
Given,sin⁡7x+6sin⁡5x+17sin⁡3x+12sin⁡xsin⁡6x+5sin⁡4x+12sin⁡2x=sin⁡7x+sin⁡5x+5sin⁡5x+5sin⁡3x+12sin⁡3x+12sin⁡xsin⁡6x+5sin⁡4x+12sin⁡2xBy using, sin⁡x+sin⁡y=2sin⁡(x+y2)cos⁡(x−y2)Therefore,=(2sin⁡(7x+5x2)cos⁡(7x−5x2)+5×(2sin⁡(5x+3x2)cos⁡(5x−3x2))+12×(2sin⁡(3x+x2)cos⁡(3x−x2))(sin⁡6x+5sin⁡4x+12sin⁡2x)=2sin⁡6xcos⁡x+10sin⁡4xcos⁡x+24sin⁡2xcos⁡x(sin⁡6x+5sin⁡4x+12sin⁡2x)=2cos⁡x(sin⁡6x+5sin⁡4x+12sin⁡2x)(sin⁡6x+5sin⁡4x+12sin⁡2x)=2cos⁡xHence, the correct option is (D).