证明三角形中,如果(sinA)^2+(sinB)^2+(sinC)^2

证明三角形中,如果(sinA)^2+(sinB)^2+(sinC)^2
证明三角形中,如果(sinA)^2+(sinB)^2+(sinC)^2

证明三角形中,如果(sinA)^2+(sinB)^2+(sinC)^2
将〈右侧变形,变为(sinA)^2+(cosA)^2+(sinB)^2+(cosB)^2,
左右就可以把(sinA)^2,(sinB)^2两项消掉,
则为
(sinc)^2

参考以下图片

证明：
∵A+B+C=180°∴C=180°-(A+B),
∴(sinA)^2+(sinB)^2+(sinB)^2=3-[(cosA)^2+(cosB)^2+(cosC)^2].
又(cosA)^2+(cosB)^2+(cosC)^2=(1+cos2A)/2+(1+cos2B)/2+(cosC)^2
=[(cos2A+cos2B)/2]+(cosC)^2+1

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证明：
∵A+B+C=180°∴C=180°-(A+B),
∴(sinA)^2+(sinB)^2+(sinB)^2=3-[(cosA)^2+(cosB)^2+(cosC)^2].
又(cosA)^2+(cosB)^2+(cosC)^2=(1+cos2A)/2+(1+cos2B)/2+(cosC)^2
=[(cos2A+cos2B)/2]+(cosC)^2+1
=cos(A+B)*cos(A-B)+[cos(A+B)]^2+1
=cos(A+B)*[cos(A+B)+cos(A-B)]+1
=-2cosAcosBcosC+1.
∴(sinA)^2+(sinB)^2+(sinC)^2=3-(-2cosAcosBcosC+1)
=2cosAcosBcosC+2.
又(sinA)^2+(sinB)^2+(sinC)^2<2
∴2cosAcosBcosC+2<2
∴cosAcosBcosC<0
即A,B,C中有且仅有一个角为钝角
∴三角形ABC为钝角三角形

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∵A+B+C=180°∴C=180°-(A+B),
∴(sinA)^2+(sinB)^2+(sinB)^2=3-[(cosA)^2+(cosB)^2+(cosC)^2].
又(cosA)^2+(cosB)^2+(cosC)^2=(1+cos2A)/2+(1+cos2B)/2+(cosC)^2
=[(cos2A+cos2B)/2]+(cosC)^2+1
=cos(A+...

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∵A+B+C=180°∴C=180°-(A+B),
∴(sinA)^2+(sinB)^2+(sinB)^2=3-[(cosA)^2+(cosB)^2+(cosC)^2].
又(cosA)^2+(cosB)^2+(cosC)^2=(1+cos2A)/2+(1+cos2B)/2+(cosC)^2
=[(cos2A+cos2B)/2]+(cosC)^2+1
=cos(A+B)*cos(A-B)+[cos(A+B)]^2+1
=cos(A+B)*[cos(A+B)+cos(A-B)]+1
=-2cosAcosBcosC+1.
∴(sinA)^2+(sinB)^2+(sinC)^2=3-(-2cosAcosBcosC+1)
=2cosAcosBcosC+2.
又(sinA)^2+(sinB)^2+(sinC)^2<2
∴2cosAcosBcosC+2<2
∴cosAcosBcosC<0
∴三角形ABC为钝角三角形

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