Question

Show that i)
(b + c)sin A/2=a COS (B-C/2)
ii)
a sin (B-C/2)=(b-c) cos A/2​

Answer 1

Answer:

Let a/sin A = b/sin B = c/sinC = k

Then,

a = k sinA, b = k sinB, c = k sinC

RHS

b-c/a cos A/2

= (ksinB - ksinC/k sinA )cos A/2

= {[2 cos B+C/2 Sin B-C/2]/ Sin A} cos A/2

=[2* Sin B-C/2 cos (pi/2-A/2) .cos A/2] / sinA

= [SinA* sinB-C/2] / sinA

= Sin B-C/2

= LHS

Hence proved

Let a/sin A = b/sin B = c/sinC = k

Then,

a = k sinA, b = k sinB, c = k sinC

RHS

b-c/a cos A/2

= (ksinB - ksinC/k sinA )cos A/2

= {[2 cos B+C/2 Sin B-C/2]/ Sin A} cos A/2

=[2* Sin B-C/2 cos (pi/2-A/2) .cos A/2] / sinA

= [SinA* sinB-C/2] / sinA

= Sin B-C/2

= LHS

Hence proved

Step-by-step explanation:

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