高一三角函数一道化简问题[siny+sinxcos(x+y)]/[cosy-sinxsin〔x+y〕]化简

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高一三角函数一道化简问题[siny+sinxcos(x+y)]/[cosy-sinxsin〔x+y〕]化简
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高一三角函数一道化简问题[siny+sinxcos(x+y)]/[cosy-sinxsin〔x+y〕]化简
高一三角函数一道化简问题
[siny+sinxcos(x+y)]/[cosy-sinxsin〔x+y〕]
化简

高一三角函数一道化简问题[siny+sinxcos(x+y)]/[cosy-sinxsin〔x+y〕]化简
siny+sinxcos(x+y)
=siny+sinx(cosxcosy-sinxsiny)
=siny+sinxcosxcosy-siny*(sinx)^2
=siny[1-(sinx)^2]+sinxcosxcosy
=sinycosx^2+sinxcosxcosy
=cosx(sinycosx+sinxcosy)
=cosxsin(x+y)
同理,计算可以得到
cosy-sinxsin(x+y)
=cosxcos(x+y)
所以
原式=tan(x+y)

分子=siny+[sin(2x+y)-siny]/2
=[sin(2x+y)+siny]/2
=sin(x+y)cosx
分母=cosy+[cos(2x+y)-cosy]/2
=[cos(2x+y)+cosx]/2
=cos(x+y)cosx
所以原式=sin(x+y)/cos(x+y)=tan(x+y)