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.