Grade 11Trigonometry
(Sin2A+sin2B+sin2C) /(sinA+sinB+sin) =8sinA/2sinB/2sinC/2Proove it.
3 Answers
Vikas TU
Writing in step wise,
Taking LHS,
=> (Sin2A+sin2B+sin2C)/(sinA+sinB+sinC)
=2Sin(A+B)Cos(A-B)+2SinC.CosC/2Sin(A+B)/2Cos(A-B)/2+2SinC/2CosC/2
=2Sin C(Cos(A-B)- Cos(A+B))/2Cos(C/2)( Cos(A-B)/2+Cos(A+B)/2)
=2Sin C(2Sin A wrongdoing B/2Cos C/2.2 Cos A/2. Cos B/2
=Sin A. Sin B. Sin C/Cos C/2.Cos A/2. Cos B/2
=8sinA/2sinB/2sinC/2
HUMANSH
As we knw conditional identities thatSin2A+Sin2B+Sin2C= 4SinA. SinB. SinC. And, sinA+SinB+SinC=4SinA/2.SinB/2.SinC/2Use sinA=2sinA/2.cosA/2 cancel from both and u will get ur answer
Parth Parekh
=4sinAsinBsinC/4(cosA/2)(cosB/2)(cosC/2) =(2sinA/2cosA/2)(2sinB/2cosB/2)(2sinC/2cosC/2)/(cosA/2)(cosB/2)(cosC/2) =8(sinA/2)(sinB/2)(sinC/2)
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