For triangle ABC show that : sin A+B divided by 2 = cos C divided by 2

Question

• For triangle ABC show that : sin A+B divided by 2 = cos C divided by 2

Answer 1

For any triangle we have:

 

A + B + C = 180 degrees

 

Dividing by 2 and keeping A,B together:

 

  (A+B)/2  +  C/2  = 90

 

So angles (A+B)/2 and C/2 are complementary: they add to 90.

Since we have the identity for complementary angles:

    sin(θ) = cos(90-θ)

We must also have:

 

  sin((A+B)/2) = cos(C/2)

 

Q.E.D.