To prove the given expression:
We start with expressing cos(A/2), cos(B/2), and cos(C/2) in terms of A, B, and C.
Given: A + B + C = pi
Using the half-angle formula for cosine:
cos(A/2) = sqrt((1 + cos A)/2)
cos(B/2) = sqrt((1 + cos B)/2)
cos(C/2) = sqrt((1 + cos C)/2)
Now, let's express cos((pi + A)/4), cos((pi - B)/4), and cos((pi + C)/4):
cos((pi + A)/4) = sqrt((1 - cos A)/2)
cos((pi - B)/4) = sqrt((1 + cos B)/2)
cos((pi + C)/4) = -sqrt((1 - cos C)/2)
Now, substituting these expressions back into the original equation:
4 cos((pi + A)/4) * cos((pi - B)/4) * cos((pi + C)/4)
= 4 * sqrt((1 - cos A)/2) * sqrt((1 + cos B)/2) * -sqrt((1 - cos C)/2)
= -4 * sqrt((1 - cos A)(1 - cos C)/4)
Now, let's look at the original expression:
cos(A/2) - cos(B/2) + cos(C/2)
Using the properties of cosine:
= sqrt((1 + cos A)/2) - sqrt((1 + cos B)/2) + sqrt((1 + cos C)/2)
Now, let's use the given property A + B + C = pi:
cos A = -(cos B + cos C)
Substituting this into the original expression:
= sqrt((1 - cos A)/2) - sqrt((1 + cos B)/2) + sqrt((1 + cos C)/2)
Now, notice that this expression is the negative of the one we derived from the right side of the equation.
So, we can rewrite the original equation as:
cos(A/2) - cos(B/2) + cos(C/2) = 4 cos((pi + A)/4) * cos((pi - B)/4) * cos((pi + C)/4)
Thus, the given expression is proven.
