在△ABC中,若2b=a+c,求证:2cos(A+C)/2=cos(A-C)/2
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在△ABC中,若2b=a+c,求证:2cos(A+C)/2=cos(A-C)/2
在△ABC中,若2b=a+c,求证:2cos(A+C)/2=cos(A-C)/2
在△ABC中,若2b=a+c,求证:2cos(A+C)/2=cos(A-C)/2
因,2b=a+c
由正弦定理得:
2sinB=sinA+sinc
2sibB=2sin(A+C)
=2*2sin(A+C)/2*con(A+C)/2
又,sinA+sinC=2sin(A+C)/2*con(A-C)/2
2*2sin(A+C)/2*con(A+C)/2=2sin(A+C)/2con(A-C)/2
故,2con(A+C)/2=con(A-C)/2
证毕.
因为,a/sinA=b/sinB=c/sinC=2R,
有,sinA=a/2R,sinB=b/2R,sinC=c/2R.
若2b=a+c,有
2sinB=sinA+sinC=2*[sin(A+C)/2*cos(A-C)/2],
2sin(B/2)*coc(B/2)=sin(A+C)/2*cos(A-C)/2,
而,A+B+C=180,B/2=90-(A+C)...
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因为,a/sinA=b/sinB=c/sinC=2R,
有,sinA=a/2R,sinB=b/2R,sinC=c/2R.
若2b=a+c,有
2sinB=sinA+sinC=2*[sin(A+C)/2*cos(A-C)/2],
2sin(B/2)*coc(B/2)=sin(A+C)/2*cos(A-C)/2,
而,A+B+C=180,B/2=90-(A+C)/2,
∴cos(B/2)=cos[90-(A+C)/2]=sin(A+C)/2.
即有,2sin(B/2)=cos(A-C)/2,.......(1)式
又因为,(A+C)/2=(180-B)/2,
cos(A+C)/2=cos(90-B/2)=sin(B/2),
2cos(A+C)/2=2sin(B/2),...........(2)式,
根据(1),(2)式,2sin(B/2)=2cos(A+C)/2,
2sin(B/2)=cos(A-C)/2,
∴2cos(A+C)/2=cos(A-C)/2,得证,等式成立.
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