】在三角形ABC中,sinA=sinBcosC/sinCcosB,判断三角形的形状.
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】在三角形ABC中,sinA=sinBcosC/sinCcosB,判断三角形的形状.
sinA +1 = (sinBcosC + sinCcosB) /sinCcosB
→sinA +1 = sin(B + C) /sinCcosB
=sin(π-A) /sinCcosB
=sinA /sinCcosB
即sinA +1 =sinA /(sinCcosB).
∵sinCcosB=(1/2)[sin(B+C)+sin(C-B)]
=(1/2)[sin(π-A)+sin(C-B)]
=(1/2)[sinA +sin(C-B)]
∴sinA +1 =sinA /{(1/2)[sinA +sin(C-B)]}
=2sinA /[sinA +sin(C-B)]
则
sin^2 A +sinAsin(C-B)+sinA +sin(C-B)=2sinA
sin^2 A +sinAsin(C-B)-sinA +sin(C-B)=0
sin(C-B)(1+sinA)=sinA(1-sinA)
sin(C-B)=sinA(1-sinA)/(1+sinA)
=sinA(1-sinA)^2/(1-sin^2 A)
=sinA(1-sinA)^2/cos^2 A
=sinA·[(1-sinA)/cosA]^2
=sinA·[tan(A/2)]^2
>0,
则角C>B
→sinA +1 = sin(B + C) /sinCcosB
=sin(π-A) /sinCcosB
=sinA /sinCcosB
即sinA +1 =sinA /(sinCcosB).
∵sinCcosB=(1/2)[sin(B+C)+sin(C-B)]
=(1/2)[sin(π-A)+sin(C-B)]
=(1/2)[sinA +sin(C-B)]
∴sinA +1 =sinA /{(1/2)[sinA +sin(C-B)]}
=2sinA /[sinA +sin(C-B)]
则
sin^2 A +sinAsin(C-B)+sinA +sin(C-B)=2sinA
sin^2 A +sinAsin(C-B)-sinA +sin(C-B)=0
sin(C-B)(1+sinA)=sinA(1-sinA)
sin(C-B)=sinA(1-sinA)/(1+sinA)
=sinA(1-sinA)^2/(1-sin^2 A)
=sinA(1-sinA)^2/cos^2 A
=sinA·[(1-sinA)/cosA]^2
=sinA·[tan(A/2)]^2
>0,
则角C>B
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