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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

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

(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...

sin(B+C)=sinBcosC+sinCcosB sin(B-C)=sinBcosC-sinCcosB cos(B+C)=cosBcosC-sinCsinB sin(B-C)=cosBcosC+sinCsinB sin(π/2-A)=cosA cos(π/2-A)=sinA

2.cos(A+B)+cosC =cos(180-c)+cosc =-cosc+cosc =0

证明: ∵A+B+C=180°, ∴sinA=sin(B+C)=sinBcosC+cosBsinC, cosA=-cos(B+C)=-(cosBcosC-sinBsinC) sin²A+2sinBsinCcosA =(sinBcosC+cosBsinC)²-2sinBsinC(cosBcosC-sinBsinC) =sin²Bcos²C+cos²Bsin²C+2...

(1)∵sinC+cosC=1?sinC2∴2sinC2cosC2+1?2sin2C2=1?sinC2∴2sinC2cosC2?2sin2C2=?sinC2∴2sin2C2?2sinC2cosC2=sinC2∴2sinC2(sin C2?cosC2)=sinC2∴sinC2?cos C2=12∴sin2C2?sinC+cos2C2=14∴sinC=34(2)由sinC2?cosC2=12>0得π4<c2<π2即π...

tanA = sinA/cosA, tanB = sinB/cosB, tanC = sinC/cosC 所以只要证明 (cosAcosBcosC)+1=sinAsinBcosC+sinBsinCcosA+sinAsinCcosB就可以了 等式右

你的公式是正确的,不过这道题给出的等式化简后是解析所讲的。你可以再仔细看看哦。2sinBcosA+2sinAcosB=4sinCcosC这个等式左右约去2,左边再用辅助角公式得到sin(A+B),然后sin(A+B)=sin(π-(A+B))=sinC,此时等式右边是2sinCcosC,然后约去sinC得...

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