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Question

If A+B+C=180o then prove the following:
(i)sin2A+sin2Bsin2C=2sinAsinBcosC
(ii) sin2A+sin2B+sin2C=2(1+cosAcosBcosC)

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Solution

We have

A+B+C=180o …… (1)

Prove that:-

(1).sin2A+sin2Bsin2C=2sinAsinBcosC

(2).sin2A+sin2Bsin2C=2(1+cosAcosBcosC)

Proof:-

Part (1).. L.H.S.

sin2A+sin2Bsin2C

=(1cos2A2)+(1cos2A2)(1cos2C)cos2θ=12sin2θ

=112(cos2A+cos2B)+cos2C1

=12(cos2A+cos2B)+cos2CcosC+cosD=2cosC+D2cosCD2

=12(2cos(2A+2B2)cos(2A2B2))+cos2C

=[cos(A+B)cos(AB)]+cos2C

=[cos(180oC)cos(AB)]+cos2Ccos(180oθ)=cosθ

=[cosCcos(AB)]+cos2C

=cosCcos(AB)+cos2C

=cosC[cos(AB)+cosC]

=cosC[cos(AB)+cos(180o(A+B))]cos(180oθ)=cosθ

=cosC[cos(AB)cos(A+B)]

=cosC(2sinAsinB)2sinAsinB=cos(AB)cos(A+B)

=2sinAsinBcosC

R.H.S.

Hence proved.

Part (2).

L.H.S.

sin2A+sin2B+sin2C

=1cos2A+sin2B+sin2C

=1(cos2Asin2B)+(1cos2C)

=2(cos(A+B)cos(AB))cos2C

=2(cos[πC]cos(AB))cos2C

=2(cosCcos(AB))cos2C

=2+cosCcos(AB)cos2C

=2+cosC(cos(AB)cosC)

=2+cosC(cos(AB)cosC)

=2+cosC(cos(AB)cos[180o(A+B)])

=2+cosC(cos(AB)cos(A+B))

=2+cosC(2cosAcosB)

=2+2cosAcosBcosC

=2(1+cosAcosBcosC)

R.H.S.

Hence proved.

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