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为什么sin(A+B)=sinC,和Cos(A+B)=%COSC

在一个三角形中,ABC的和为180度

∵A、B、C是三角形的三个内角∴A+B=π-C对于A,cos(A+B)=cos(π-C)=-cosC,故A错对于B,sin(A+B)=sin(π-C)=sinC,故B对对于C,tan(A+B)=tan(π-C)=-tanC,故C错对于D,sin A+B 2 = cos C 2 ,故D错故选B

cos(a+b+c) =cos[(a+b)+c] =cos(a+b)cosc-sin(a+b)sinc =(cosacosb-sinasinb)cosc-(sinacosb+cosasinb)sinc =cosacosbcosc-sinasinbcosc-sinacosbsinc-cosasinbsinc

由正弦定理,a:b:c=sinA:sinB:sinC,所以由acosA+bcosB=ccosC得sinAcosA+sinBcosB=sinCcosC,所以sin(2A)+sin(2B)=sin(2C)和差化积,2sin(A+B)cos(A-B)=sin(2C)=2sinCcosC,所以cos(A-B)=cosC因为A,B,C都是三角形的内角,所以A-B=C...

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-...

解: 2cosC(acosB+bcosA=c 由正弦定理得: 2cosC(sinAcosB+sinBcosA)=sinC 2cosCsin(A+B)-sinC=0 2cosCsinC-sinC=0 sinC(2cosC-1)=0 C为...

(本小题满分12分)(Ⅰ)∵ sinB+sinC sinA = 2-cosB-cosC cosA ∴sinBcosA+sinCcosA=2sinA-cosBsinA-cosCsinA∴sinBcosA+cosBsinA+sinCcosA+cosCsinA=2sinAsin(A+B)+sin(A+C)=2sinA…(3分)sinC+sinB=2sinA…(5分)所以b+c=2a…(6分)(Ⅱ)由...

(1)因为A+B+C=π,所以B=π-(A+C),代入sinAcosC+3sinAsinC-sinB-sinC=0的,sinAcosC+3sinAsinC-sin(A+C)-sinC=0,sinAcosC+3sinAsinC-sinAcosC-cosAsinC-sinC=03sinAsinC-cosAsinC-sinC=0,因为sinC≠0,所以3sinA-cosA=1,即sin(A?π6)=12...

解: cosA=3/5>0,A为锐角 sinA=√(1-cos²A)=√[1-(3/5)²]=4/5>√2/2 π/4

(1) ∵cosA=2/3,∴sinA=√(1-cos²A)=√5/3 ∵sinB=√5cosC sinB=sin(A+C)=sinAcosC+cosAsinC ∴sinAcosC+cosAsinC=√5cosC ∴√5/3cosC+2/3sinC=√5cosC ∴ sinC=√5cosC ,∴tanC=√5 (2)若a=√2,∵ sinA=√5/3 ∴2R=a/sinA=√2/(√5/3)=3√10/5 ∵ sinC=√5cosC, ...

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