Question

Y equal to root of 1 - sin 2x by 1 + sin 2x prove that dy by dx +

Answer 1

Note cos(π/4 - x) = cos(π/4 ) * cos(x) + sin(π/4 ) * sin(x)

 

                            = (cos(x) + sin(x))/√(2)

 

cos2(π/4 - x) = (cos(x) + sin(x))2/2

                     = (cos2(x) + sin2(X) + 2*cos(x)*sin(x))/2

                     = (1 + sin(2x))/2

 

From this sec2(π/4 - x) = 2/(1 + sin(2x))

 

Also, 1 - sin(2x) = sin2(x) + cos2(x) - 2*sin(x)*cos(x)

                          = (cos(x) - sin(x))2

 

Similarly, 1 + sin(2x) = (cos(x) + sin(x))2

 

Also (1 - sin(2x))/(1 + sin(2x)) = ((cos(x) - sin(x))/(cos(x) + sin(x)))2

                                                = ((cos(x) - sin(x))*(cos(x) + sin(x))/(cos(x) + sin(x))2)2

                                                = ((cos2(x) - sin2(x))/(1 + sin(2x)))2

                                                = (cos(2x)/(1 + sin(2x)))2

 

Now y =√((1 - sin(2x))/(1 + sin(2x)))

 

          = √((cos(2x)/(1 + sin(2x)))2)

 

          = cos(2x)/(1 + sin(2x))

 

Take the derivative:

 

   y' = -2sin(2x)/(1 + sin(2x)) + cos(2x) (-1)(2cos(2x))/(1 + sin(2x))2

 

       = -2/(1 + sin(2x))2*(sin(2x)*(1 + sin(2x)) + cos2(2x))

 

       = -2/(1 + sin(2x))2*(sin(2x) + sin2(2x) + cos2(2x))

 

       = -2/(1 + sin(2x))2*(1 + sin(2x))

 

       = -2/(1 + sin(2x))

 

      = -sec2(π/4 - x), from the identity above