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In any triangle ABC, show that sin B−C/2=b−c/acos A/2

Arya J , 7 Years ago
Grade 11
anser 1 Answers
Aditya Gupta
note that if we prove 2sinA/2sin (B−C)/2=b−c/a 2cos A/2*sinA/2, then by dividing both sides by sinA/2 we will have the eqiuvalent result.
but 2sinA/2sin (B−C)/2= cos(A+C – B)/2 – cos(A+B – C)/2= cos(90-B) – cos(90 – C)= sinB – sinC
and 2cos A/2*sinA/2= sinA
so, we have to prove
sinB – sinC= (b – c)/a * sinA
or sinB/sinA – sinC/sinA= (b – c)/a , which directly follows from the sine rules!
QED
Last Activity: 7 Years ago
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