Question

In a triangle show that b^2sin2c+c^2sin2b=2bcsina

Answer 1

 
b2sin2C + c2sin2B 
 
= b2(2sinC cosC) + c2(2sinB cosB)
Since, sinC/c =sinB/b = sinA/a = R
 
So, 
= b2(2Rc cosC) + c2(2RbcosB)
 
=b2[2Rc ×{b2+a2 - c2}/(2ab)]+c2[2Rb×{a2 + c2-b2}/(2ac)]
 
=2Rbc [ {b2+a2 - c2}/(2a) + {a2 + c2-b2}/(2a)]
 
=2Rbc [ a]
 
= 2bc(Ra)
 
=2bcsinA