Question

prove that cos7x-cos8x/1+2cos5x=cos2x-cos3x

Answer 1

L.H.S = (cos7x – cos8x) / 1 + 2cos5x

      (Adding cos2x and -cos2x in the numerator)

       = (cos7x + cos2x) – (cos8x + cos2x) / 1 + 2cos5x

     (Using cosA + cosB = 2cos(A+B/2)cos(A-B/2) )

       = (cos7x + cos 2x) – (2cos5x cos3x) / 1 + 2cos5x

     (Adding cos3x and -cos3x in the numerator)

        = (cos7x + cos2x + cos 3x) – (2cos5x cos3x + cos3x) / 1 + 2cos5x

       = (2cos5x cos2x + cos2x) – cos3x (2cos5x + 1) / 2cos 5x + 1

       = cos2x (2cos5x + 1) – cos3x (2cos5x + 1) / 2cos5x + 1

      = (2cos5x + 1) (cos2x – cos3x) / ( 2cos5x + 1)

       = cos2x – cos3x = R.H.S

Answer 2

Step-by-step explanation:

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