在三角行ABC中,sinA+sinC=2sinB,A-C=π/3,求sinB的值要完整过程.

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在三角行ABC中,sinA+sinC=2sinB,A-C=π/3,求sinB的值
要完整过程.

sinA+sinC=2sin[(A+C)/2]×cos[(A-C)/2]=2sin[(π-B)/2]×cos(π/6)=2sin(π/2-B/2)×√3/2=√3cos(B/2)=√3×(1+cosB)/2.
已知sinA+sinC=2sinB,
则:√3×(1+cosB)/2=2sinB
化简得:cosB=4√3sinB/3-1,将等式两边平方:
则:1-sin²B=(4√3/3sinB-1)²
解得:sinB=8√3/19.

sinA+sinC=2sinB,
和差化积公式,
2sin[(A+C)/2]cos[(A-C)/2]=4sin(B/2)cos(B/2),
sin[(π-B)/2]*cos[(A-C)/2]=2sin(B/2)cos(B/2),
cos(B/2)*cos[(A-C)/2]=2sin(B/2)cos(B/2),......(1)
cos(π/6)=2sin(B...

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sinA+sinC=2sinB,
和差化积公式,
2sin[(A+C)/2]cos[(A-C)/2]=4sin(B/2)cos(B/2),
sin[(π-B)/2]*cos[(A-C)/2]=2sin(B/2)cos(B/2),
cos(B/2)*cos[(A-C)/2]=2sin(B/2)cos(B/2),......(1)
cos(π/6)=2sin(B/2),.....(2)
sin(B/2)=√3/4,
cos(B/2)=√(1-3/16)=√13/4,
∴sinB=2sin(B/2)cosB/2)
=2*(√3/4)*√13/4.
=√39/8.

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