Question

For any triangle ABC , prove that. a+b/c = cos (A-B/2)/sin C/2

Answer 1

Answer:

Step-by-step explanation:

For any triangle ABC with sides a, b, c opposite to A, B, C angles,

a = 2RsinA

b = 2RsinB

c = 2RsinC

where R is the circum-radius of the triangle ABC

(a+b)/c = (sinA + sinB)/sinC

= ( 2sin( (A+B)/2 )cos( (A-B)/2 ) ) / sinC

A + B + C = 180

(A+B)/2 = 90 - C/2

= ( 2sin(90 - C/2)cos( (A-B)/2 ) ) / 2sin(C/2)cos(C/2)

= ( cos(C/2)cos( (A-B)/2 ) ) / sin(C/2)cos(C/2)

= cos( (A-B)/2 ) / sin(C/2)