在ABC中 COS2C=-九分之一 求sinc
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cos2A+cos2B=2cos(A+B)cos(A-B)1+cos2C=2(cosC)^2cos(A+B)=-cosC-cosCcos(A-B)=(cosC)^2所以cosC=0或-cos(A-B)
(1)cos2C=1-2(sinC)^2=-1/9sinC=√5/3(2)2sinC等于根五倍的sinA,得2c=√5a,c=√5,cos=2/34+b^2-2*2bcosC=5,b=3b=3,c=√
(1)因为cos2C=cos(C+C)=(cosC)平方-(sinC)平方=-3/4(cosC)平方+(sinC)平方=1得(sinC)平方=7/8所以,sinC=根号14/4(2)
4cos²C/2-cos2C=7/22(1+cosC)-2cos²C+1=7/24cos²C-4cosC+1=0(2cosC-1)²=0cosC=1/2C=60
sinC=sin(π-(A+B))=sin(A+B)cos2C=cos2(π-(A+B))=cos2(A+B)∴sinA(sinB+cosB)-sin(A+B)=0sinAsinB+sinAcosB)
cos2A+cos2B+cos2Ccos2A+cos2B+cos2C=(cos2A+cos2B)+(cos2B+cos2C)+(cos2A+cos2C).用和差化积公式cos(a)+cos(b)=2c
∵由sinA(sinB+cosB)-sinC=0∴sinAsinB+sinAcosB-sin(A+B)=0.∴sinAsinB+sinAcosB-sinAcosB-cosAsinB=0.∴sinB(s
当a=2,2sinA=sinC时,由正弦定理asinA=csinC,得:c=4由cos2C=2cos2C-1=-14,及0<C<π得cosC=±64由余弦定理c2=a2+b2-2abcosC,得b2±
A+C=120度m*n=-根号3sin2A-cos2C=-根号3sin2(120-C)-cos2C=根号3sin(60-2C)-cos2C=1/2cos2C-根号3/2sin2C=-sin(2C-π/
由正弦定理得a/sinA=b/sinB,因为acosA=bcosB,所以sinAcosB-cosAsinB=sin(A-B)=0,所以∠A=∠B.cos2A+cos2B-cos2C=2cos2A-co
cos2A+cos2B=2cos(A+B)cos(A-B)1+cos2C=2(cosC)^2cos(A+B)=-cosC-cosCcos(A-B)=(cosC)^2所以cosC=0或-cos(A-B)
超简单~的思路一般都是联系条件(三角形)那么就把“大角化小角,一步一步慢慢走咯(柴)”.1-2sin^2A+1-2sin^2B-1+2sin^2C=1得sin^2A+sin^2B=sin^2C在用上正
sin²2C+sin2C×sinC+cos2C=1,4sin²C*cos²C+2sin²CcosC+1-2sin²C=1,2cos²C+co
4sin^2(A+B)-cos2C=2[1-cos(A+B)]-(2cos^2C-1)=7/2,4cos^2C-4cosC+1=0cosC=1/2,C=60度a+b=5,a^2+2ab+b^2=25,
这属于多变量的极值问题,可以采取所谓的“冻结变量法”.显然A,B,C三个角中至少有两个锐角,不妨假设C为锐角,固定角C不变,由和差化积公式:cos2A+cos2B=2cos(A+B)cos(A-B)=
A=45B=120C=15sinC=sin(a+b)带入第一个公试可以得出A=45再把第二个公式中的sinB=sin(A+C)带掉然后cos2C=cosC的平方-sinC的平方然后配平再化简后得到si
4sin²(A+B/2)-cos2C=2-2cos(A+B)-(2cos²C-1)=3+2cosC-2cos²C=7/2所以cosC=1/2,所以∠C=60°cosC=a
/>4sin²[(A+B)/2-cos2C=7/2,2(1-cos(A+B))-(2cos²C-1)=7/2,2(1+cosC)-(2cos²C-1)=7/2,2+2co
sinA*(sinB+cosB)-cosC=0sinA*(sinB+cosB)+cos(A+B)=0sinA*(sinB+cosB)+cosAcosB-sinAsinB=0sinAcosB+cosAc
根据正弦定理,(a+b)/a=sinB/(sinB-sinA)=(sinA+sinB)/sinA∴sinA·sinB=(sinB+sinA)(sinB-sinA)=2sin[(B+A)/2]·cos[