What is cosx*sinx if cosx-sinx=1/3?

We have to find cos x*sin x given that cos x - sin x =
1/3

cos x - sin x = 1/3

square
both the sides

=> (cos x - sin x)^2 =
1/9

=> (cos x)^2 + (sin x)^2 - 2*sin x* cos x =
1/9

use (cos x)^2 + (sin x)^2 =
1

=> 1 - 2*sin x* cos x =
1/9

=> 2*sin x * cos x = 1 -
1/9

=> 2*sin x * cos x =
8/9

=> sin x* cos x =
4/9

The required value of sin x* cos x =
4/9