SOLUTION: If A, B, C are angles of a triangle, prove that cosA + cosB + cosC = 4sin(A/2)sin(B/2)sin(C/2)

Algebra ->  Triangles -> SOLUTION: If A, B, C are angles of a triangle, prove that cosA + cosB + cosC = 4sin(A/2)sin(B/2)sin(C/2)      Log On


   

Question 1131925: If A, B, C are angles of a triangle, prove that
cosA + cosB + cosC = 4sin(A/2)sin(B/2)sin(C/2)
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
As A,B and C are angles of a triangle, we have A%2BB%2BC=180
and
A%2BB=180-C
or
%28A%2BB%29%2F2=90-C%2F2

Hence

cos%28A%29%2Bcos%28B%29%2Bcos%28C%29

=2cos%28A%2BB%29%2F2%29%2Acos%28A-B%29%2F2%29+%2B1-2sin%5E2%28C%2F2%29-1.......since %28A%2BB%29%2F2=90-C%2F2

=2cos%2890-C%2F2%29%2Acos%28%28A-B%29%2F2%29%2B1+-2sin%5E2%28C%2F2%29...since cos%2890-C%2F2%29=sin%28C%2F2%29

=2sin%28C%2F2%29%2Acos%28%28A-B%29%2F2%29%2B1+-2sin%5E2%28C%2F2%29......factor

=2sin%28C%2F2%29%28cos%28%28A-B%29%2F2%29-sin%28C%2F2%29%29%2B1 ........from sin%28%28A%2BB%29%2F2%29=sine%2890-C%2F2%29=>sine%28%28A%2BB%29%2F2%29=sin%28C%2F2%29

=2sin%28C%2F2%29%28cos%28%28A-B%29%2F2%29-sin%28%28A%2BB%29%2F2%29%29%2B1

=2sin%28C%2F2%29%282sin%28A%2F2%29%2Asin%28B%2F2%29%29%2B1

=1%2B4sin%28A%2F2%29%2Asin%28B%2F2%29sin%28C%2F2%29