Establish the identity

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

Hint:

sin 2A = 2sinAcosA

treat x/2 as A

To establish the identity, we need to simplify the left-hand side (LHS) and the right-hand side (RHS) of the equation separately, and then show that they are equal.

Starting with the LHS:
(cosx/2 - sinx/2)^2

To simplify this expression, we can use the identity:
(a - b)^2 = a^2 - 2ab + b^2

Applying this identity to our expression:
(cosx/2 - sinx/2)^2 = (cosx/2)^2 - 2(cosx/2)(sinx/2) + (sinx/2)^2

Expanding this:
= cos^2(x/2)/4 - 2 cos(x/2) sin(x/2)/4 + sin^2(x/2)/4

Now, let's consider the RHS:
1 - sinx

To simplify this expression, we just leave it as it is.

Now, let's equate the LHS and RHS, and see if we can make them equal.

(cos^2(x/2) - cos(x/2)sin(x/2) + sin^2(x/2))/4 = 1 - sinx

Multiplying through by 4 to get rid of the denominator:
cos^2(x/2) - 2cos(x/2)sin(x/2) + sin^2(x/2) = 4 - 4sinx

Now, let's use the following trigonometric identity:
cos^2(x/2) + sin^2(x/2) = 1

Replacing this in our equation:
1 - 2cos(x/2)sin(x/2) = 4 - 4sinx

Simplifying further:
2cos(x/2)sin(x/2) = 3 - 4sinx

Now, we can divide both sides of the equation by 2:
cos(x/2)sin(x/2) = (3 - 4sinx)/2

We can also write the RHS of the equation as:
cos(x/2)sin(x/2) = (3/2) - 2sinx

Now, it is clear that the LHS and RHS are not equal, indicating that the given equation is not an identity.