Question

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

Answer 1

By using sine formula

a/sinA=b/sinB=c/sinC=2R

thus,a=2RsinA;b=2RsinB;c=2RsinC

RHS

(b-c/a)cosA/2=(2RsinB-2RsinC/2RsinA)cosA/2

=2sin(B-C/2)cos(B+C/2)/2sinA/2cosA/2 x cos A/2

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

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

=sin(B-C/2)

RHS

Answer 2

Answer:

here is your answer mate

Step-by-step explanation:

By using sine formula

a/sinA=b/sinB=c/sinC=2R

thus,a=2RsinA;b=2RsinB;c=2RsinC

RHS

(b-c/a)cosA/2=(2RsinB-2RsinC/2RsinA)cosA/2

=2sin(B-C/2)cos(B+C/2)/2sinA/2cosA/2 x cos A/2

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

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

=sin(B-C/2)

RHS