If A+B+C=180, then \frac{\sin 2A+\sin 2B+\sin 2C}{\cos A+\cos B+\cos C−1}=8

Autohelmvt

Autohelmvt

Answered question

2022-01-23

If A+B+C=180, then sin2A+sin2B+sin2CcosA+cosB+cosC1=8cosA2cosB2cosC2
Then 2A+2B+2C=360
So
sin2C=sin(2A+2B)
Putting that in the equation
2sin(A+B)sin(AB)2sin(A+B)cos(A+B)cosA+cosBcos(A+B)+1
2sin(A+B)[sin(AB)cos(A+B)]cosA+cosBcos(A+B)+1
I don’t know how to proceed. Please help me continue

Answer & Explanation

Alfred Mueller

Alfred Mueller

Beginner2022-01-24Added 10 answers

We will proceed by simplifying the numerator and denominator separately and repeatedly use some well-known formulae.
For simplifying the numerator:
sin2A+sin2B+sin2C=2sin(A+B)cos(AB)+2sinCcosC
=2sin(πC)cos(AB)+2sinCcosC
=2sinC[cos(AB)+cos(π(A+B))]
=2sinC[cos(AB)cos(A+B)]
=4sinAsinBsinC
For simplifying the denominator:
cosA+cosB+cosC1=2cos(A+B2)cos(AB2)2sin2C2
=2cos(π2C2)cos(AB2)2sin2C2
=2sinC2[cos(AB2)sinC2]
=2sinC2[cos(AB2)cos(A+B2)]
=4sinA2sinB2sinC2

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