verify identity

(sinx-cosx)^2=1-sin(2x)

LS = sin^2 x - 2sinxcosx + cos^2 x

= 1 - sin (2x)
= RS

To verify the identity (sinx-cosx)^2 = 1 - sin(2x), we will simplify the left-hand side and the right-hand side of the equation separately and show that they are equal.

Let's start with the left-hand side:

(sin x - cos x)^2

Now, we will expand the square using the formula (a - b)^2 = a^2 - 2ab + b^2:

(sin x)^2 - 2(sin x)(cos x) + (cos x)^2

Using the trigonometric identity sin^2(x) + cos^2(x) = 1, we can rewrite the equation as:

1 - 2(sin x)(cos x) + 1

Simplifying further, we have:

2 - 2(sin x)(cos x)

Now, let's move on to the right-hand side:

1 - sin(2x)

Using the double-angle formula for sine, which states that sin(2x) = 2(sin x)(cos x), we can substitute this value into our equation:

1 - 2(sin x)(cos x)

We can see that the right-hand side is the same as the simplified left-hand side, which means the identity is verified.

Therefore, we have verified the identity (sin x - cos x)^2 = 1 - sin(2x).