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now if we use  sin^2 x=(1-cos2x)/2

then we are not getting solution like this

 

please check

 

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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

 

edited by

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