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

							
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
						

							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.SIntegrate it and get answer as[sin2x]/2 - [sin3x]/3Derivative of it would be 3sin3x - 2sin2x
						

							Q.1) (cos7x – cos8x) / 1 + 2cos5x      Integrate it and get answer as[sin2x]/2 - [sin3x]/3Derivative of it would be 3sin3x - 2sin2x