a+b = 3pi/2

==> cos(a+b) = cos(3pi/2)

==> cos(a+b) = 0

prove that:

sin2a - sin2b = 0

We know that:

sina - sinb = 2cos(a+b)/2 *sin(a-b)/2

Then,

sin2a - sin2b = 2*cos(2a+2b)/2 *sin(2a-2b)/2

= 2*cos(a+b) *sin(a-b)

= 2*0*sin(a-b) = 0

==> **sin2a - sin2b = 0**

a+b = 3pi/2 To prove sin2a -sin2b = 0

As a +b = 3pi/2,

Solution:

We use sinA-sinB = 2cos(A+B)/2 * sin(A-B)/2

Therefore sin2a - sin2b = 2 cos (2a+2b)/2 *sin(2a-2b)

= 2cos (a+b)*cos(a-b)

= 2cos(pi/2)*sin(a-b)

= 0, as cos(3pi/2) = 0.

We'll transform the difference of the sine functions into a product.

sin 2a - sin 2b = 2cos [(2a+2b)/2]*sin[(2a-2b)/2]

We'll factorize by 2 and we'll get:

sin 2a - sin 2b = 2cos [2(a+b)/2]*sin [2(a-b)/2]

sin 2a - sin 2b = 2cos (a+b)*sin (a-b)

We'll substitute the sum a+b by the value 3pi/2.

sin 2a - sin 2b = 2cos (3pi/2)*sin(a-b)

We know that cos (3pi/2) = 0

Since a factor from the product is zero, that means that the entire product is also zero.

**sin 2a - sin 2b = 0 q.e.d.**