若函数f(tanx)=sin2x+sinxcosx+1,求f(x)的解析式由于上课这题用换元法回答错误,老师让我用换元法写一遍,答案是f(x)=(2x²+x+1)/(x²+1)
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![若函数f(tanx)=sin2x+sinxcosx+1,求f(x)的解析式由于上课这题用换元法回答错误,老师让我用换元法写一遍,答案是f(x)=(2x²+x+1)/(x²+1)](/uploads/image/z/12511807-7-7.jpg?t=%E8%8B%A5%E5%87%BD%E6%95%B0f%28tanx%29%3Dsin2x%2Bsinxcosx%2B1%2C%E6%B1%82f%EF%BC%88x%EF%BC%89%E7%9A%84%E8%A7%A3%E6%9E%90%E5%BC%8F%E7%94%B1%E4%BA%8E%E4%B8%8A%E8%AF%BE%E8%BF%99%E9%A2%98%E7%94%A8%E6%8D%A2%E5%85%83%E6%B3%95%E5%9B%9E%E7%AD%94%E9%94%99%E8%AF%AF%2C%E8%80%81%E5%B8%88%E8%AE%A9%E6%88%91%E7%94%A8%E6%8D%A2%E5%85%83%E6%B3%95%E5%86%99%E4%B8%80%E9%81%8D%2C%E7%AD%94%E6%A1%88%E6%98%AFf%28x%29%3D%EF%BC%882x%26%23178%3B%2Bx%2B1%EF%BC%89%2F%28x%26%23178%3B%2B1%29)
若函数f(tanx)=sin2x+sinxcosx+1,求f(x)的解析式由于上课这题用换元法回答错误,老师让我用换元法写一遍,答案是f(x)=(2x²+x+1)/(x²+1)
若函数f(tanx)=sin2x+sinxcosx+1,求f(x)的解析式
由于上课这题用换元法回答错误,老师让我用换元法写一遍,答案是f(x)=(2x²+x+1)/(x²+1)
若函数f(tanx)=sin2x+sinxcosx+1,求f(x)的解析式由于上课这题用换元法回答错误,老师让我用换元法写一遍,答案是f(x)=(2x²+x+1)/(x²+1)
答:
设t=tanx,sinx=tcosx
代入(sinx)^2+(cosx)^2=1有:
(t^2+1)(cosx)^2=1
(cosx)^2=1/(1+t^2)
所以:
f(tanx)=(sinx)^2+sinxcosx+1
f(t)=t^2/(1+t^2)+t(cosx)^2+1
=t^2/(1+t^2)+t/(1+t^2)+1
=(t^2+t+1+t^2)/(1+t^2)
所以:
f(x)=(2x²+x+1)/(x²+1)