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为什么3CosBCosC+3sinBsinClosBCosC=%1能变成3(CosBCosC%s...

您好!【问题】我是想知道3cosBcosC-6cosBcosC+3sinBsinC不是应该等于-3sinBsinC+3cosBcosC吗?【解析】假设他们相等.则: 3cosBcosC-6cosBcosC+3sinBsinC=-3sinBsinC+3cosBcosC ∴ -6cosBcosC+3sinBsinC=-3sinBsinC ∴ 6

3cosBcosC+3sinBsinC-6cosBcosC=3sinBsinC-3cosBcosC=-3(cosBcosC-sinbsinC)=-3cos(B+C)

3cosBcosC+1=3sinBsinC+cos2A3cosBcosC-3sinBsinC=cos2A-13(cosBcosC-sinBsinC)=cos2A-13cos(B+C) = cos2A-1-3cosA = 2cosA-1-12cosA+3cosA-2=0(cosA+2)(2cosA-1)=02cosA-1=0cosA=1/2A=60°a=2√3B+C=120°b=asinB/

(Ⅰ)由3cosBcosC+1=3sinBsinC+cos2A,得3(cosBcosC-sinBsinC)=cos2A-1,即3cos(B+C)=2cos2A-2,即2cos2A+3cosA-2=0…(3分)可得:(2cosA-1)(cosA+2)=0,可得:cosA=12或cosA=-2(舍去),可得:A=π3…6分(II)由S=12bcsinA=34bc=53,得bc=20.又b=5,所以c=4.-----(8分)由余弦定理,得a2=b2+c2-2bccosA=25+16-20=21,故a=21.---(10分)又由正弦定理,得sinBsinC=bca2sin2A=2021*34=57.----(12分)

三角形ABC的内角A,B,C的对边分别为a,b,c,已知△ABC的面积为a^2/(3sinA).且6cosBcosC=1,a=3,求△ABC的周长.解:△ABC的面积为a^2/(3sinA)=(1/2)bcsinA,由正弦定理,sinBsinC=2/3,①6cosBcosC=1,cosBcosC=1/6,② ②-①得cos(B+

把右边的式子全部移到左边,把3cosBcosC-3sinBsinC合并成3根号2cos(B+C),即负3根号2cosA所以总式子变成了-3根号2cosA-2cosA的平方+2=0,解方程即可

三角形ABC各角对应边abc且a=b+c+√3ab 1求角A 2设a=√3,S为三角形面积,求S+ 3cosBcosC的最大值,并指出此时B的值.(1)由a=b+c+√3ab ,b+c-a=-√3bccosA=(b+c-a)/2bc=-√3/2,A在第二象限,∴∠A=150

cosA=cos(180-B-C)=-cos(B+C)=-(cosBcosC-sinBsinC)=-(3sinBsinC-sinBsinC)=-2sinBsinCcos(B-C)=cosBcosC+sinBsinC=3sinBsinC+sinBsinC=4sinBsinCcosA/cos(B-C) =-2sinBsinC/4sinBsinC=-1/2

(cosA-3cosC)sinB=(3sinC-sinA)cosBcosAsinB-3cosCsinB=3sinCcosB-sinAcosBcosAsinB+sinAcosB=3cosCsinB+3sinCcosBsin(A+B)=3sin(B+C)sinC=3sinA展开后移项,用和角公式.最后一步是用到三角形中公式:sin(A+B)=sinC.

解: 6cosbcosc =6X1/2 [cos(b+c)+cos(b-c)] =3[cos(b+c)+cos(b-c)] =3cos(b+c)+3cos(b-c)答:6cosbcosc可以变成3cos(b+c)+3cos(b-c) .

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