函数f(x)=sin(x+π/3)-根号3cos(x+π?3),x∈(0,2π)闭区间,的单调递增递减区间是多少f(x)=sin(x+π/3)-根号3cos(x+π?3),x∈(0,2π)这个的化简最要命的,老师不要跳步,
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![函数f(x)=sin(x+π/3)-根号3cos(x+π?3),x∈(0,2π)闭区间,的单调递增递减区间是多少f(x)=sin(x+π/3)-根号3cos(x+π?3),x∈(0,2π)这个的化简最要命的,老师不要跳步,](/uploads/image/z/5177081-65-1.jpg?t=%E5%87%BD%E6%95%B0f%28x%29%3Dsin%28x%2B%CF%80%2F3%EF%BC%89-%E6%A0%B9%E5%8F%B73cos%EF%BC%88x%2B%CF%80%3F3%EF%BC%89%2Cx%E2%88%88%EF%BC%880%2C2%CF%80%EF%BC%89%E9%97%AD%E5%8C%BA%E9%97%B4%2C%E7%9A%84%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E9%80%92%E5%87%8F%E5%8C%BA%E9%97%B4%E6%98%AF%E5%A4%9A%E5%B0%91f%28x%29%3Dsin%28x%2B%CF%80%2F3%EF%BC%89-%E6%A0%B9%E5%8F%B73cos%EF%BC%88x%2B%CF%80%3F3%EF%BC%89%2Cx%E2%88%88%EF%BC%880%2C2%CF%80%EF%BC%89%E8%BF%99%E4%B8%AA%E7%9A%84%E5%8C%96%E7%AE%80%E6%9C%80%E8%A6%81%E5%91%BD%E7%9A%84%2C%E8%80%81%E5%B8%88%E4%B8%8D%E8%A6%81%E8%B7%B3%E6%AD%A5%2C)
函数f(x)=sin(x+π/3)-根号3cos(x+π?3),x∈(0,2π)闭区间,的单调递增递减区间是多少f(x)=sin(x+π/3)-根号3cos(x+π?3),x∈(0,2π)这个的化简最要命的,老师不要跳步,
函数f(x)=sin(x+π/3)-根号3cos(x+π?3),x∈(0,2π)闭区间,的单调递增递减区间是多少
f(x)=sin(x+π/3)-根号3cos(x+π?3),x∈(0,2π)
这个的化简最要命的,老师不要跳步,
函数f(x)=sin(x+π/3)-根号3cos(x+π?3),x∈(0,2π)闭区间,的单调递增递减区间是多少f(x)=sin(x+π/3)-根号3cos(x+π?3),x∈(0,2π)这个的化简最要命的,老师不要跳步,
本题类型叫做用引入辅助解的方法,将一个角的两个函数化为一个角的一个三角函数.
f(x)=sin(x+π/3)-√3cos(x+π/3)
=2[(1/2)*sin(x+π/3)-(√3/2)*cos(x+π/3)]
=2[cos(π/3)*sin(x+π/3)-sin(π/3)*cos(x+π/3)]
=2sin[(x+π/3)-(π/3)]
=2sinx
当x=π/2 时,函数达到最大值2
当x=3π/2时,函数达到最小值-2
因此函数2sinx在[0 , 2π)上的单调性是先增后减再增,是一个大N字样的图像,所以f(x)的单调递减
区间是:
[π/2 , 3π/2]
按你的要求整个过程已细化到最大体积,若有不明白的地方,欢迎追问.