cho tam giác ABC chứng minh rằng:
cosA+cosB-cosC= \(4cos\frac{A}{2}.cos\frac{B}{2}.sin\frac{C}{2}-1\)
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30 tháng 4 2016 lúc 12:57
cho A , B , C là 3 góc của tam giác ABC . chứng minh rằng : a) sin2A + sin2B + sin2C = 4sinAsinBsinC ; b) cosA + cosB + cosC = 1 = 4sin\(\frac{A}{2}\)sin\(\frac{B}{2}\)sin\(\frac{C}{2}\) ; c) cos2A + cos2B + cos2C = 1 - 2cosAcosBcosC
cho tam giác abc có 3 góc nhọn. Vẽ đường cáo AD, BE, CF cắt nhau tại H. Chứng minh:
a) \(0< cos^2A+cos^2B+cos^2C< 1\)
b)\(2< sin^2A+sin^2B+sin^2C< 3\)
c)sinA + sinB + sinC < 2( cosA + cosB + cosC)
d)sinB . cosC + sinC . cosB = sinA
e)tanA + tanB + tanC = tanA . tanB . tanC
Cho tam giác ABC. CMR:
a) sinA + sinB + sinC = 4cos(A/2)cos(B/2)cos(C/2)
b) cosA + cosB + cosC = 1 + 4sin(A/2)sin(B/2)sin(C/2)
c) sin2A + sin2B + sin2C = 4sinA.sinB.sinC
d) cos2A + cos2B + cos2C = -(1 + 4cosA.cosB.cosC)
Cho tam giác ABC. CMR:
a) sinA + sinB + sinC = 4cos(A/2)cos(B/2)cos(C/2)
b) cosA + cosB + cosC = 1 + 4sin(A/2)sin(B/2)sin(C/2)
c) sin2A + sin2B + sin2C = 4sinA.sinB.sinC
d) cos2A + cos2B + cos2C = -(1 + 4cosA.cosB.cosC)
Cho tam giác ABC. CMR:
a) sinA + sinB + sinC = 4cos(A/2)cos(B/2)cos(C/2)
b) cosA + cosB + cosC = 1 + 4sin(A/2)sin(B/2)sin(C/2)
c) sin2A + sin2B + sin2C = 4sinA.sinB.sinC
d) cos2A + cos2B + cos2C = -(1 + 4cosA.cosB.cosC)
Cho tam giác ABC. CMR:
a) sinA + sinB + sinC = 4cos(A/2)cos(B/2)cos(C/2)
b) cosA + cosB + cosC = 1 + 4sin(A/2)sin(B/2)sin(C/2)
c) sin2A + sin2B + sin2C = 4sinA.sinB.sinC
d) cos2A + cos2B + cos2C = -(1 + 4cosA.cosB.cosC)
chứng minh rằng
trong ΔABC ta có
cosA + cosB + cosC = 1+ 4sin\(\frac{A}{2}\)sin\(\frac{B}{2}\)sin\(\frac{C}{2}\)
cho tam giác nhọn ABC.
Cm: \(cosA+cosB+cosC=1+4sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}\)
cho tam giác ABC . chứng minh:
a, sin(A+B)sinC. ; cos (A+B)cos-C; tan ( A+B) -tan C
b, sinfrac{A+B}{2}cosfrac{C}{2} ; cosfrac{A+B}{2}sinfrac{C}{2} ; tanfrac{A+B}{2}cotfrac{C}{2}
c, tan A+tanB+tanC tanA.tanB.tanc( tam giác không vuông)
d, sinA+sinB+sinC 4cosfrac{A}{2}cosfrac{B}{2}cosfrac{C}{2}
e, cos A+cosB+cosC 1+4sinfrac{A}{2}sinfrac{B}{2}sinfrac{C}{2}
f, sin2A+sin2B+sin2C 4sinAsinBsinC
g, cos 2A+cos2B+cos2C1-2cosAcosBcosC
Đọc tiếp
cho tam giác ABC . chứng minh:
a, sin(A+B)=sinC. ; cos (A+B)=cos-C; tan ( A+B)= -tan C
b, \(sin\frac{A+B}{2}=cos\frac{C}{2}\) ; \(cos\frac{A+B}{2}=sin\frac{C}{2}\) ; tan\(\frac{A+B}{2}=cot\frac{C}{2}\)
c, tan A+tanB+tanC= tanA.tanB.tanc( tam giác không vuông)
d, sinA+sinB+sinC= \(4cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}\)
e, cos A+cosB+cosC= \(1+4sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}\)
f, sin2A+sin2B+sin2C= 4sinAsinBsinC
g, cos 2A+cos2B+cos2C=1-2cosAcosBcosC
\(A+B+C=180^0\Rightarrow A+B=180^0-C\)
\(\Rightarrow sin\left(A+B\right)=sin\left(180^0-C\right)=sinC\)
\(cos\left(A+B\right)=cos\left(180^0-C\right)=-cosC\)
\(tan\left(A+B\right)=tan\left(180^0-C\right)=-tanC\)
b/ \(\frac{A+B+C}{2}=90^0\Rightarrow\frac{A+B}{2}=90^0-\frac{C}{2}\)
\(\Rightarrow sin\frac{A+B}{2}=sin\left(90^0-\frac{C}{2}\right)=cos\frac{C}{2}\)
\(cos\frac{A+B}{2}=cos\left(90^0-\frac{C}{2}\right)=sin\frac{C}{2}\)
\(tan\frac{A+B}{2}=tan\left(90-\frac{C}{2}\right)=cot\frac{C}{2}\)
c/ \(A+B=180^0-C\Rightarrow tan\left(A+B\right)=-tanC\)
\(\Leftrightarrow\frac{tanA+tanB}{1-tanA.tanB}=-tanC\)
\(\Leftrightarrow tanA+tanB=-tanC+tanA.tanB.tanC\)
\(\Leftrightarrow tanA+tanB+tanC=tanA.tanB.tanC\)
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d/ \(sinA+sinB+sinC=2sin\frac{A+B}{2}cos\frac{A-B}{2}+2sin\frac{C}{2}.cos\frac{C}{2}\)
\(=2cos\frac{C}{2}.cos\frac{A-B}{2}+2sin\frac{C}{2}.cos\frac{C}{2}\)
\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+sin\frac{C}{2}\right)\)
\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+cos\frac{A+B}{2}\right)\)
\(=4cos\frac{C}{2}.cos\frac{A}{2}.cos\frac{B}{2}\)
e/
\(cosA+cosB+cosC=2cos\frac{A+B}{2}cos\frac{A-B}{2}+1-2sin^2\frac{C}{2}\)
\(=1+2sin\frac{C}{2}.cos\frac{A-B}{2}-2sin^2\frac{C}{2}\)
\(=1+2sin\frac{C}{2}\left(cos\frac{A-B}{2}-sin\frac{C}{2}\right)\)
\(=1+2sin\frac{C}{2}\left(cos\frac{A-B}{2}-cos\frac{A+B}{2}\right)\)
\(=1+4sin\frac{C}{2}.sin\frac{A}{2}sin\frac{B}{2}\)
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f/
\(sin2A+sin2B+sin2C=2sin\left(A+B\right).cos\left(A-B\right)+2sinC.cosC\)
\(=2sinC.cos\left(A-B\right)+2sinC.cosC\)
\(=2sinC\left(cos\left(A-B\right)+cosC\right)\)
\(=2sinC\left[cos\left(A-B\right)-cos\left(A+B\right)\right]\)
\(=4sinC.sinA.sinB\)
g/
\(cos^2A+cos^2B+cos^2C=\frac{1}{2}+\frac{1}{2}cos2A+\frac{1}{2}+\frac{1}{2}cos2B+cos^2C\)
\(=1+\frac{1}{2}\left(cos2A+cos2B\right)+cos^2C\)
\(=1+cos\left(A+B\right).cos\left(A-B\right)+cos^2C\)
\(=1-cosC.cos\left(A-B\right)+cos^2C\)
\(=1-cosC\left(cos\left(A-B\right)-cosC\right)\)
\(=1-cosC\left[cos\left(A-B\right)+cos\left(A+B\right)\right]\)
\(=1-2cosC.cosA.cosB\)
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