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已知A,B,C为锐角,满足Cos^2A+Cos^2B+Cos^2C+2CosA...

将其看成cosC的一元二次方程,则可以写成cosC+2cosAcosBcosC+cosA+cosB-1=0.因此cosC=-cosAcosB±√(cosAcosB-cosA-cosB+1)=-cosAcosB±√[(1-cosA)(1-cosB)]=-cosAcosB±√(sinAsinB)=-cosAcosB±|

第2个 提示 SEC^2a+sec^2b+sec2c 用2次柯西可以解出 第一个 tan9+1/tan9-tan27-1/tan27=2/sin18-2/sin54=2(sin54-sin18)/sin54sin18=2(sin54-sin18)cos18/sin18sin54

不妨设C为钝角2cosA=1+cos2A,2cosB=1+cos2B则t=1+(cos2A+cos2B+2cosC)/2=1+[2cos(A+B)cos(A-B)+2cosC]/2=1+cosC[cosC-cos(A-B)]因cosC<0且|A-B|<90°,则cosC-cos(A-B)<0则t=1+cosC[cosC-cos(A-B)]>1显然cos(A-B)≥1,又cosC<0则cosC[cosC-cos(A-B)]≤cosC(cosC-1)则t≤1+cosC(cosC-1)=(cosC-1/2)+3/4t<(-1-1/2)+3/4=3(因0>cosC>-1)故1<t<3

cos^2A=cos^2(B+C)=1-sin^2(B+C)sin(B+C)=sinBcosC+sinCcosB所以cos^2A+cos^2B+cos^2C=cos^2B+cos^2C-(sin^2Bcos^2C+cos^2Bsin^2C)-2(sinBcosCcosBsinC) +1=1所以cos^2B+cos^2C-(sin^2Bcos^2C+cos^2Bsin^2C)=2(

2a,2b,2c是空间向量角

sinB=c/,△abc是等腰三角形,或,或;sinC*sin(B+2C)因为;2sinC/2cosC/2cos(2B+C)/,B+C=π/sinB*sin(2B+C)=c/,或;2cosC/2]=-sin(2B+C)sinC同样cos^2(B+C)-cos^C=-sin(B+2C)sinB所以bsin(2B+C)sinC=csin(B+2C)sinBb/2*(-sin(2B+C)/,

你的式子有一项好像抄错了如果原题是求证a(cos2B-cos2C)+b(cos2C-cos2A)+c(cos2A-cos2B)=0的话证明如下:a(cos2B-cos2C)+b(cos2C-cos2A)+c(cos2A-cos2B)=a(1-2sinB-1+2sinC)+b(1-2sinC-1+2sinA)+c(1-2sin

由cos^2A=sin^2B+cos^2C-sinAsinB,得1-sin^2A=sin^2B+(1-sin^2C)-sinAsinBsin^2A+sin^2B-sin^2C=sinAsinB由正弦定理,得a^2+b^2-c^2=ab 由余弦定理,得cosC=(a^2+b^2-c^2)/2ab=1/2因为0° 评论0 0 0

2*2RsinA*2RsinB=(sin^2A+sin^2B-sin^2C)/2sinAsinB,同样;12=(7-2v6)/12;由已知;2sinCsinA,cosB=(sin^2C+sin^2A-sin^2B)/:cosC=(a^2+b^2-c^2)/2ab=4R^2(sin^2A+sin^2B-sin^2C)/2sinBsinC,cosA=1/2,》sinB=sin(A+C)=sinAcosC+

(1)∵(cosA)^2-(cosB)^2=cos(π/6-A)cos(π/6+A),∴(cosA)^2-(cosB)^2=(1/2)cos(π/3)+(1/2)cos2A∴(cosA)^2-(cosB)^2=1/4+(1/2)[2(cosA)^2-1]=1/4+(cosA)^2-1/2,∴

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