若2sin(θ+π/3)=1+cos(θ+π/3),则tanθ/2=_____
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若2sin(θ+π/3)=1+cos(θ+π/3),则tanθ/2=_____
若2sin(θ+π/3)=1+cos(θ+π/3),则tanθ/2=_____
若2sin(θ+π/3)=1+cos(θ+π/3),则tanθ/2=_____
2sin(θ+π/3)=1+cos(θ+π/3)
sin(θ+π/3)=(1+cos(θ+π/3))/2 降幂公式逆用
2sin(θ/2+π/6)cos(θ/2+π/6)=[ cos(θ/2+π/6) ]^2 正弦倍角公式
2sin(θ/2+π/6)=cos(θ/2+π/6)
tan(θ/2+π/6)=1/2
( tanθ/2+tanπ/6)/(1-tanθ/2 * tanπ/6) =1/2
tanθ/2 =(3-2根号3)/(6+根号3） 结果你在再算算,思想,方法就这样
由题意可得：
2sin(θ+π/3)=1+cos(θ+π/3)=cos0+cos(θ+π/3)=2cos[(θ+π/3)/2]cos[(θ+π/3)/2]
所以sin(θ+π/3)=cos[(θ+π/3)/2]cos[(θ+π/3)/2]
2sin[(θ+π/3)/2]cos[(θ+π/3)/2]=cos[(θ+π/3)/2]cos[(...
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由题意可得：
2sin(θ+π/3)=1+cos(θ+π/3)=cos0+cos(θ+π/3)=2cos[(θ+π/3)/2]cos[(θ+π/3)/2]
所以sin(θ+π/3)=cos[(θ+π/3)/2]cos[(θ+π/3)/2]
2sin[(θ+π/3)/2]cos[(θ+π/3)/2]=cos[(θ+π/3)/2]cos[(θ+π/3)/2]
因为cos(θ+π/3)不等于0
所以2sin[(θ+π/3)/2]=cos[(θ+π/3)/2]
所以tan(θ/2+π/6)=1/2
所以[tan(θ/2)+√3/3]/[1-√3/3tan(θ/2)]=1/2
解得：tanθ/2=(15-8√3)/33
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