已知:sinA+sinB+sinC=0 cosA+cosB+cosC=0 计算下式:sin²A+sin
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已知:sinA+sinB+sinC=0 cosA+cosB+cosC=0 计算下式:sin²A+sin²B+sin²C
![已知:sinA+sinB+sinC=0 cosA+cosB+cosC=0 计算下式:sin²A+sin](/uploads/image/z/8448167-47-7.jpg?t=%E5%B7%B2%E7%9F%A5%EF%BC%9AsinA%2BsinB%2BsinC%3D0%E3%80%80cosA%2BcosB%2BcosC%3D0+%E8%AE%A1%E7%AE%97%E4%B8%8B%E5%BC%8F%EF%BC%9Asin%26%23178%3BA%2Bsin%26%2317)
sinA+sinB+sinC=0
cosA+cosB+cosC=0
sinA+sinB=-sinC
cosA+cosB=-cosC
平方得
(sinA+sinB)^2=(sinC)^2 (1)
(cosA+cosB)^2=(cosC)^2 (2)
两式相加得
1+2sinAsinB+2cosAcosB+1=1
所以
2sinAsinB+2cosAcosB=-1
cos(A-B)=-1/2
(2)-(1)
(cosC)^2-(sinC)^2=(cosA+cosB)^2-(sinA+sinB)^2
cos2C=cos2A+cos2B+2cos(A+B)
sin²A+sin²B+sin²C
=3-( cos²A+cos²B+cos²C)
=3-3/2=3/2
再问: cos²A+cos²B+cos²C为何为3/2?
再答: cos²A+cos²B+cos²C =(1+cos2A)/2+(1+cos2B)/2+(1+cos2C)/2 =3/2+(cos2A+cos2B+cos2C)/2 =3/2+cos2A+cos2B+cos(A+B) =3/2+cos((A+B)+(A-B))+cos((A+B)-(A-B))+cos(A+B) =3/2+2cos(A+B)cos(A-B)+cos(A+B) =3/2
再问: 可否用cos(A-B)=-1/2 得出A-B=±120°,算得A-B=B-C=C-A=±120°? 直接计算sin²A+sin²B+sin²C=3/2
cosA+cosB+cosC=0
sinA+sinB=-sinC
cosA+cosB=-cosC
平方得
(sinA+sinB)^2=(sinC)^2 (1)
(cosA+cosB)^2=(cosC)^2 (2)
两式相加得
1+2sinAsinB+2cosAcosB+1=1
所以
2sinAsinB+2cosAcosB=-1
cos(A-B)=-1/2
(2)-(1)
(cosC)^2-(sinC)^2=(cosA+cosB)^2-(sinA+sinB)^2
cos2C=cos2A+cos2B+2cos(A+B)
sin²A+sin²B+sin²C
=3-( cos²A+cos²B+cos²C)
=3-3/2=3/2
再问: cos²A+cos²B+cos²C为何为3/2?
再答: cos²A+cos²B+cos²C =(1+cos2A)/2+(1+cos2B)/2+(1+cos2C)/2 =3/2+(cos2A+cos2B+cos2C)/2 =3/2+cos2A+cos2B+cos(A+B) =3/2+cos((A+B)+(A-B))+cos((A+B)-(A-B))+cos(A+B) =3/2+2cos(A+B)cos(A-B)+cos(A+B) =3/2
再问: 可否用cos(A-B)=-1/2 得出A-B=±120°,算得A-B=B-C=C-A=±120°? 直接计算sin²A+sin²B+sin²C=3/2
已知:sinA+sinB+sinC=0 cosA+cosB+cosC=0 计算下式:sin²A+sin
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