Question

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

Answer 1

Step-by-step explanation:

In a triangle ABC

a. =b. =c. =2R

sinA. sinB. sinC

a=2R sinA;b=2RsinB;c=2R sinC

R. H. S

(b-c)/a×cosA/2

(2RsinB-2R sinC)/2R sinA×cosA/2

sinB-sinC×cosA/2

sin A

sinB-sinC ×cosA/2

sin(180-(B+C))

sinB-sinC×cos(90-(B+C)/2)

sin(B+C)

2 cos((B+C)/2)sin((B-C)/2)×sin((B+C)/2)

2 sin((B+C)/2) cos((B+C)/2)

sin((B-C)/2)=L. H. S

hence proved

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