Question

in ∆ABC,prove that sin(A-B/2)=(a-b/c)cos(c/2)​

Answer 1

Answer:

Step-by-step explanation:

The sum of all angles in a triangle is 180°

Therefore,

A + B + C = 180°

=> A + B = 180° - C

=> (A + B)/2 = (180° - C)/2 = 90° - C/2

In the above question,

LHS = cos((A + B)/2)

substituting (A + B)/2 with 90°- C/2

= cos(90° - C/2)

cos(90° - θ) = sinθ

Therefore,

= sin(C/2) which is also = RHS

LHS = RHS

Hence proved.