SOLUTION: If A+B+C=π then establish the given relation (sin2A+sin2B+sin2C)/(4cosA/2 cosB/2 cosC/2)=8 sinA/2.sinB/2.sinC/2

Algebra ->  Trigonometry-basics -> SOLUTION: If A+B+C=π then establish the given relation (sin2A+sin2B+sin2C)/(4cosA/2 cosB/2 cosC/2)=8 sinA/2.sinB/2.sinC/2      Log On


   

Question 1014707: If A+B+C=π then establish the given relation
(sin2A+sin2B+sin2C)/(4cosA/2 cosB/2 cosC/2)=8 sinA/2.sinB/2.sinC/2
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
Prove first that
sin2A + sin2B + sin2C = 4sinA*sinB*sinC.
sin2A + sin2B + sin2C
=sin2A+%2B+sin2B+%2B+sin%282%2Api-2%28A%2BB%29%29
=sin2A+%2B+sin2B+-+sin2%28A%2BB%29
=sin2A+%2B+sin2B+-+sin%282A%2B2B%29
=sin2A+%2B+sin2B+-+sin2A%2Acos2B+-+cos2A%2Asin2B
= sin2A(1 - cos2B) + sin2B(1 - cos2A)
=2sin2A%2A%28sinB%29%5E2+%2B+2sin2B%2A%28sinA%29%5E2
=4sinA%2AcosB%2A%28sinB%29%5E2+%2B+4sinB%2AcosB%2A%28sinA%29%5E2
=4sinAsinB(cosAsinB+sinAcosB)
=4sinAsinBsin(A+B)
=4sinAsinBsin%28pi-C%29
=4sinAsinBsinC
Since %284sinA%2AsinB%2AsinC%29%2F%284cos%28A%2F2%29%2Acos%28B%2F2%29%2Acos%28C%2F2%29%29
=,
the main result follows.