Sin^2A+Sin^2B+Sin^2C = 2+2CosACosBCosC When A+B+C=180

Abdur1234567890 Abdur1234567890

• 03.04.2019
• Math
• Secondary School

answered Sin^2A+Sin^2B+Sin^2C = 2+2CosACosBCosC when A+B+C=180°​ See answers rahman786khalilu rahman786khalilu

Given A+B+C = 180

C = 180 -( A+B)

cos C = cos 180 - (A +B) = - cos (A+B)

now, Sin^2A + 1 - Cos^2B + Sin^2C

1 -(Cos^2B - Sin^2A) + Sin ^2C

1 - Cos (A+B) Cos (A - B) + 1- Cos^2C

2 + CosC [Cos(A -B) + Cos ( A + B)]

2 + 2 Cos A CosB CosC

Hence L.H.S = R.H.S

may this will help us mark as brainliest

pradeeppadhy pradeeppadhy

Answer:

sin^2A+sin^2B+sin^2C-2cosAxosBcosC

= 1-cos^2A+1-cos^2B+1-cos^2C-2cosAcosBcosC

= 3-cos^2A-cos^2B-cos^2C-2cosAcosBcosc

=1/2(6-2cos^2A-2cos^2B-2cos^2C-4cosAcosB

cos C

=then do it yourself I am going to sleeping

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