搜狗做题网高一三角函数一道化简问题
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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
[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)
=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)