www.fltk.net > 在△ABC中,A,B,C分别是内角A,B,C的对边,且(2A%C)CosBosC=0.(1)求...

在△ABC中,A,B,C分别是内角A,B,C的对边,且(2A%C)CosBosC=0.(1)求...

(1)由(2a-c)cosB-bcosC=0及正弦定理得,(2sinA-sinC)cosB-sinBcosC=0,即2sinAcosB-sin(B+C)=0,因为A+B+C=π,所以sin(B+C)=sinA,因为sinA≠0,所以cosB=1 2 ,由B是三角形内角得,B=π 3 ,(2)由(1)得,B=π 3 ,则f(x)=-2cos(2x+B)=-2

(1)2cos(B+C)+cos2A=-3/2-2cosA+cos2A=-3/24cosA-4cosA+1=0cosA=1/2A=60°(2)a=3b+c=3cosA=1/2a=b+c-2bccosAb+c+2bc=9b=1,c=2或b=2,c=1

由条件及正弦定理得sinBcosC+sinCcosB=-2sinAcosB.即sin(B+C)=-2sinAcosB.∵A+B+C=π,A>0∴sin(B+C)=sinA,又sinA≠0,∴cosB=-12,而B∈(0,π),∴B=2π3.故选:C.

(1)(2a+c)cosB+bcosC=0由正弦定理:(2sinA+sinC)cosB+sinBcosC=02sinAcosB+cosBsinC+sinBcosC=02sinAcosB+sin(B+C)=02sinAcosB+sinA=0cosB=-1/2、B=2π/3.(2)b=√13、a+c=4.(a+c)^2=a^2+c^2+2ac=16、a^2+c^2=16-2ac.

1.B=2π/32.y=sin(A+C)-2sinAsinC=sinB-2sinAsin(π/3-A)=3/4-√3cosA+sinA=2sin(A-π/6)+3/4A∈(0,π/3)∴A-π/6∈(-π/6,π/6)2sin(A-π/6)+3/4∈(3/4,5/4)∴y∈(3/4,5/4)

(I)由已知得cosBcosC=b2a+c,由正弦定理得cosBcosC=sinB2sinA+sinC.即2sinAcosB+sinCcosB=-sinBcosC,即2sinAcosB+sin(B+C)=0.…3分∵B+C=π-A,∴sin(B+C)=sin(π-A)=sinA,∴cosB=12,∴B=2π

(1)2cos(B-C)=4sinBsinC-12cosBcosC+2sinBsinC=4sinBsinC-12cosBcosC-2sinBsinC=-12cos(B+C)=-1cos(B+C)=-1/2cosA=(π-B-C)=-cos(B+C)=1/2A=π/3,即60°(2)sinA=√3/2(sinB)/2=1/3 sinB=2/3a/sinA=b/sinBb=asinB/sin

(Ⅰ)由cos(B-C)-2sinBsinC=-12,得cosBcosC-sinBsinC=-12,即cos(B+C)=-12.∴B+C=2π3,故A=π3. (Ⅱ)由sinB2=13,得cosB2=223,∴sinB=2sinB2cosB2=429. ∵bsinB=asinA,∴b429=332,解得b=869.

2cos(A+2C)=1-4sinBsinC2cos[(A+C)+C]=1-4sinBsinC2cos[π-(B-C)]=1-4sinBsinC-2cos(B-C)=1-4sinBsinC-2(cosBcosC+sinBsinC)=1-4sinBsinC-2(cosBcosC-sinBsinC)=1cos(B+C)=-1/2cos(π-A)=-1/2cosA=1/2A=60°1.a/SinA=b/sinB

bcosC=(2a-c)cosBb* a 2 + b 2 - c 2 2ab =(2a-c)* a 2 + c 2 - b 2 2ac a 2 + b 2 - c 2 2a = a 2 + c 2 - b 2 c - a 2 + c 2 - b 2 2a 2a 2 2a = a 2 + c 2 - b 2 c a 2 + c 2 - b 2

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