SO here we need TO DO comparison between ,
sin2(ax+b) / 2a + c1 = -cos(2(ax+b)) / 4a + c2
taking constants for LHS = c1 and RHS = c2 , and we know that a and b are constants with respect to the given equation.
L.H.S = sin2(ax+b) / 2a + c1
= 2sin2(ax+b) / 4a + c1 ( dividing and multiplying equation by 2 )
= 2 (1- cos2(ax+b)) / 4a + c1 ( since SIN2X = 1- COS2X)
= 2 - 2cos2(ax+b) / 4a + c1
=(1 + (1 - 2cos2(ax+b) ) / 4a ) + (c1 )
= (1/ 4a) + (1- 2cos2(ax+b) )/ 4a ) + (c1 )
=(1- 2cos2(ax+b) )/ 4a ) + (c1 + 1/ 4a)
= - cos2(ax+b) / 4a + ( c3)
(as cos 2x = -(1 -2cos2x) and let c1 + 1/4a = c3)
assuming c3 and c2 to be equal ( approximately equal) as constant.
= R.H.S.
therefore , L.H.S = R.H.S.
cos2x= cos2x-sin2x = cos2x - (1-cos2x) = -1 + 2cos2x = -(1-2cos2x)
so, (1-2cos2x) = -cos2x