已知函数f(x)=2倍根号3sinxcosx+2cos²x-1(x∈R) (1)求函数的最小正周期(2)若f(x)=6/5,接着上边的x∈[π/4,π/2],cos2x的值
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/30 00:57:00
![已知函数f(x)=2倍根号3sinxcosx+2cos²x-1(x∈R) (1)求函数的最小正周期(2)若f(x)=6/5,接着上边的x∈[π/4,π/2],cos2x的值](/uploads/image/z/989015-23-5.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D2%E5%80%8D%E6%A0%B9%E5%8F%B73sinxcosx%2B2cos%26sup2%3Bx-1%28x%E2%88%88R%29+%281%29%E6%B1%82%E5%87%BD%E6%95%B0%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%EF%BC%882%EF%BC%89%E8%8B%A5f%EF%BC%88x%EF%BC%89%3D6%2F5%2C%E6%8E%A5%E7%9D%80%E4%B8%8A%E8%BE%B9%E7%9A%84x%E2%88%88%5B%CF%80%2F4%2C%CF%80%2F2%5D%2Ccos2x%E7%9A%84%E5%80%BC)
已知函数f(x)=2倍根号3sinxcosx+2cos²x-1(x∈R) (1)求函数的最小正周期(2)若f(x)=6/5,接着上边的x∈[π/4,π/2],cos2x的值
已知函数f(x)=2倍根号3sinxcosx+2cos²x-1(x∈R) (1)求函数的最小正周期(2)若f(x)=6/5,
接着上边的x∈[π/4,π/2],cos2x的值
已知函数f(x)=2倍根号3sinxcosx+2cos²x-1(x∈R) (1)求函数的最小正周期(2)若f(x)=6/5,接着上边的x∈[π/4,π/2],cos2x的值
(1)
f(x)=2√3sinxcosx+2cos²x-1=√3sin2x+cos2x=2(√3/2sin2x+1/2cos2x)=2sin(2x+π/6)
函数的最小正周期T=π
(2)
f(x)=2sin(2x+π/6)=6/5
sin(2x+π/6)=3/5
x∈[π/4,π/2],
cos(2x+π/6)=-4/5
cos2x=cos(2x+π/6-π/6)=cos(2x+π/6)cosπ/6+sin(2x+π/6)sinπ/6=-4/5*√3/2+3/5*1/2=3/10-2√3/5
先整理原式,f(x)=根号3sin2x+cos2x=2sin(2x+π/6)
(1)所以周期就是2π/2=π
(2)设y=cos2x,因为根号3sin2x+cos2x=6/5,所以根号3sin2x=6/5-cos2x
两边平方3(1-y*y)=36/25+y*y-12y/5整理的100y*y-60y-39=0
然后由于x∈[π/4,π/2],所以2x∈[π/2,π]...
全部展开
先整理原式,f(x)=根号3sin2x+cos2x=2sin(2x+π/6)
(1)所以周期就是2π/2=π
(2)设y=cos2x,因为根号3sin2x+cos2x=6/5,所以根号3sin2x=6/5-cos2x
两边平方3(1-y*y)=36/25+y*y-12y/5整理的100y*y-60y-39=0
然后由于x∈[π/4,π/2],所以2x∈[π/2,π],所以cos2x<0
然后解方程得y=(3-4根号3)/10
收起