# if A+B+C = 180° prove that : sin²A + sin²B + sin²C = 2 + 2cosA.cosB.sinC

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### 1 Answer

`sin^2A+sin^2B+sin^2C = 2+2cosA.cosB.cosC`

I can write,

`sin^2A` as,

`sin^2A = (1-cos(2A))/2`

Therefore,

LHS =

`= (1-cos(2A))/2+(1-cos(2B))/2+(1-cos(2C))/2`

`=3/2 - (cos(2A)+cos(2B)+cos(2C))`

`= 1/2(3 - (2cos(A+B)cos(A-B)+cos(2C)))`

`C = 180 - (A+B)`

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

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

Therefore,

LHS

`=3/2 - (-2cos(C)cos(A-B)+cos(2C))`

`cos(2C) = 2cos^2(C) - 1`

LHS

`=1/2(3 - (-2cos(C)cos(A-B)+2cos^2(C)-1))`

`= 1/2(4 - (2cos(C)(cos(C)-cos(A-B)))`

`= 1/2(4 - 2cos(C)(-cos(A+B)-cos(A-B)))`

`= 1/2(4+2cos(C)(cos(A+B)cos(A-B)))`

`= 1/2(4+2cos(C)xx2cos(A)cos(B))`

`= 2 + 2cos(A)cos(B)cos(C)`

Therefore, if `A+B+C = 180,`

`sin^2A+sin^2B+sin^2C = 2+2cosA.cosB.cosC`

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