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
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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) (Show Source): You can put this solution on YOUR website!
Prove first that
sin2A + sin2B + sin2C = 4sinA*sinB*sinC.
sin2A + sin2B + sin2C
=
=
=
=
= sin2A(1 - cos2B) + sin2B(1 - cos2A)
=
=
=4sinAsinB(cosAsinB+sinAcosB)
=4sinAsinBsin(A+B)
=
=4sinAsinBsinC
Since
=,
the main result follows.
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