Question

三角形中，(sinA)^2+(sinB）^2+(sinC)^2<2,求证此三角形为钝角三角形

Answer 1

根据正弦定理，a/sinA=b/sinB=c/sinC,所以，（sinA）^2=a^2*(sinB）^2/b^2,同理，（sinC）^2=c^2*(sinB）^2/b^2，所以，(sinA)^2+(sinB）^2+(sinC)^2=（a^2+b^2+c^2)/(b^2*(sinB）^2)<2,sinB<=1,说明（a^2+b^2+c^2)/b^2<2,说明a^2+c^2<b^2,所以它是钝角三角形

Answer 2

同一三角形中sinA=sin(B+C) 则有(sinA)^2+(sinB）^2+(sinC)^2<2 [sin(B+C)]^2+(sinB）^2+(sinC)^2<2 (sinBcosC+cosBsinC)^2+(sinB）^2+(sinC)^2<2 (sinB)^2(cosC)^2+(cosB)^2(sinC)^2+2sinBcosCcosBsinC+(sinB）^2+(sinC)^2<2 (sinB)^2(cosC)^2+(cosB)^2(sinC)^2+2sinBcosCcosBsinC+1-(cosB）^2+1-(cosC)^2<2 [(sinB)^2-1](cosC)^2+(cosB)^2[(sinC)^2-1]+2sinBcosCcosBsinC+2<2 -(cosB)^2(cosC)^2-(cosB)^2(cosC)^2+2sinBcosCcosBsinC+2<2 -2(cosB)^2(cosC)^2+2sinBcosCcosBsinC+2<2 -(cosB)^2(cosC)^2+sinBcosCcosBsinC<0 cosBcosC(sinBsinC-cosBcosC)<0,因为cosA=-cos(B+C) 所以cosBcosCcosA<0 可推出A,B,C中必有一个角为钝角,则三角形ABC为钝角三角形

Answer 3

证明：有三角形内恒等式：
(sinA)^2+(sinB)^2+(sinC)^2=2cosAcosBcosC+2,
所以，
cosAcosBcosC<0，为钝角三角形。

Answer 4

同一三角形中sina=sin(b+c)
则有(sina)^2+(sinb）^2+(sinc)^2<2
[sin(b+c)]^2+(sinb）^2+(sinc)^2<2
(sinbcosc+cosbsinc)^2+(sinb）^2+(sinc)^2<2
(sinb)^2(cosc)^2+(cosb)^2(sinc)^2+2sinbcosccosbsinc+(sinb）^2+(sinc)^2<2
(sinb)^2(cosc)^2+(cosb)^2(sinc)^2+2sinbcosccosbsinc+1-(cosb）^2+1-(cosc)^2<2
[(sinb)^2-1](cosc)^2+(cosb)^2[(sinc)^2-1]+2sinbcosccosbsinc+2<2
-(cosb)^2(cosc)^2-(cosb)^2(cosc)^2+2sinbcosccosbsinc+2<2
-2(cosb)^2(cosc)^2+2sinbcosccosbsinc+2<2
-(cosb)^2(cosc)^2+sinbcosccosbsinc<0
cosbcosc(sinbsinc-cosbcosc)<0,因为cosa=-cos(b+c)
所以cosbcosccosa<0
可推出a,b,c中必有一个角为钝角,则三角形abc为钝角三角形

Answer 5

三角形中sinA=sin(B+C)
则有(sinA)^2+(sinB）^2+(sinC)^2<2
[sin(B+C)]^2+(sinB）^2+(sinC)^2<2
(sinBcosC+cosBsinC)^2+(sinB）^2+(sinC)^2<2
(sinB)^2(cosC)^2+(cosB)^2(sinC)^2+2sinBcosCcosBsinC+(sinB）^2+(sinC)^2<2
(sinB)^2(cosC)^2+(cosB)^2(sinC)^2+2sinBcosCcosBsinC+1-(cosB）^2+1-(cosC)^2<2
[(sinB)^2-1](cosC)^2+(cosB)^2[(sinC)^2-1]+2sinBcosCcosBsinC+2<2
-(cosB)^2(cosC)^2-(cosB)^2(cosC)^2+2sinBcosCcosBsinC+2<2
-2(cosB)^2(cosC)^2+2sinBcosCcosBsinC+2<2
-(cosB)^2(cosC)^2+sinBcosCcosBsinC<0
cosBcosC(sinBsinC-cosBcosC)<0,
因为cosA=-cos(B+C)
所以cosBcosCcosA<0
可推出A,B,C中必有一个角为钝角,
则三角形ABC为钝角三角形
根据正弦定理，a/sinA=b/sinB=c/sinC,
所以，（sinA）^2=a^2*(sinB）^2/b^2,
同理，（sinC）^2=c^2*(sinB）^2/b^2，
所以，(sinA)^2+(sinB）^2+(sinC)^2=（a^2+b^2+c^2)/[b^2*(sinB）^2]<2,sinB<=1,
说明（a^2+b^2+c^2)/b^2<2,
说明a^2+c^2
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