sinA+sinC=2sin[(A+C)/2]×cos[(A-C)/2]怎么推
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sinA+sinC=2sin[(A+C)/2]×cos[(A-C)/2]怎么推
在三角形ABC中,SinA+SinC=2SinB,A-C=π\3,求SinB的值?
在三角形ABC中,SinA+SinC=2SinB,A-C=π\3,求SinB的值?
设(A+C)/2=x,(A-C)/2=y,则A=x+y,B=x-y,sinA+sinB =sin(x+y)+sin(x-y) =sinxcosy+cosxsiny+sinxcosy-cosxsiny =2sinxcosy =2sin[(A+C)/2]×cos[(A-C)/2] sinA+sinC=2sin[(A+C)/2]×cos[(A-C)/2]=2cosBcosπ\6=根号(3)cosB=2SinB tanB=根号(3)/2,则cosB>0,故而1+{根号(3)/2}^2=7/4=cos^(-2)B,即cosB=2/根号(7) sinB=tanB*cosB=根号(3/7)
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cos^2A - cos^2B + sin^2C=2cosA *sinB *sinC证明
cos^2A - cos^2B + sin^2C=2cosA *sinB *sinC
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