# What is the value of the product sinx*cosx if the difference sinx-cosx is 5/7?

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Given that sinx - cosx = 5/7

We need to know that value sinx*cosx

Let us square both sides.

==> (sinx - cosx)^2 = (5/7)^2

==> sin^2 x - 2sinx*cosx + cos^2 x = 25/49

But we know that sin^2cx + cos^2 x = 1

==> -2sinx*cosx + 1 = 25/49

We will subtract 1 from both sides.

==> -2sinx*cosx = 25/49 - 1

==> -2sinxcosx = -24/49

Now we will divide by -2

==> sinx*cosx = -24/49*-2 = 12/49

**Then the value is sinx*cosx = 12/49**

We have sin x - cos x = 5/7 and we need to find (sin x)*(cos x).

sin x - cos x = 5/7

square both the sides

(sin x)^2 + (cos x)^2 - 2*(sin x)*(cos x) = 25/49

it is known that (sin x)^2 + (cos x)^2 = 1

=> 1 - 2*(sin x)*(cos x) = 25/49

=> 2*(sin x)*(cos x) = 1 - 25/49

=> 2*(sin x)*(cos x) = 24/49

=> (sin x)*(cos x) = 12/49

**The value of (sin x)*(cos x) = 12/49**

We'll raise the given difference to square:

(sin x - cos x)^2 = (5/7)^2

We'll expand the square:

(sin x)^2 - 2sin x*cos x + (cos x)^2 = 25/49

We'll apply the Pythagorean identity:

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

1 - 2sin x*cos x = 25/49

2sin x*cos x = 1 - 25/49

2sin x*cos x = 24/49

sin x*cos x = 12/49

**The requested product is: sin x*cos x = 12/49.**