prove that:
sinC+sinD =2sin((C+D)/2)*cos((C-D)/2)
take a look here:
http://math.ucsd.edu/~wgarner/math4c/textbook/chapter6/product_sum_formulas.htm
To prove the equation sinC + sinD = 2sin((C+D)/2)cos((C-D)/2), we can use the trigonometric identity for the sum of angles:
sin(A + B) = sinA * cosB + cosA * sinB
Let's start with the left-hand side of the equation:
sinC + sinD
Using the identity, we can write:
= sin(C + D/2 + D/2)
Now, let's focus on the right-hand side of the equation:
2sin((C+D)/2)cos((C-D)/2)
Using the double angle identity for sine, sin(2x) = 2sin(x)cos(x), we can rewrite it as:
2 * sin((C+D)/2) * cos((C-D)/2)
Now, let's compare both sides:
sin(C + D/2 + D/2) = 2 * sin((C+D)/2) * cos((C-D)/2)
By comparing the two expressions, we can see that they are equal. Therefore, we have proved the equation:
sinC + sinD = 2sin((C+D)/2)cos((C-D)/2)