已知函数f(x)=2√ 3sinxcosx+2cos^2x-1若f(Xo)=6/5,Xo属于【π/4,π/2】,求cos2Xo的值
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![已知函数f(x)=2√ 3sinxcosx+2cos^2x-1若f(Xo)=6/5,Xo属于【π/4,π/2】,求cos2Xo的值](/uploads/image/z/1989132-60-2.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D2%E2%88%9A+3sinxcosx%2B2cos%5E2x-1%E8%8B%A5f%EF%BC%88Xo%EF%BC%89%3D6%2F5%2CXo%E5%B1%9E%E4%BA%8E%E3%80%90%CF%80%2F4%2C%CF%80%2F2%E3%80%91%2C%E6%B1%82cos2Xo%E7%9A%84%E5%80%BC)
已知函数f(x)=2√ 3sinxcosx+2cos^2x-1若f(Xo)=6/5,Xo属于【π/4,π/2】,求cos2Xo的值
已知函数f(x)=2√ 3sinxcosx+2cos^2x-1
若f(Xo)=6/5,Xo属于【π/4,π/2】,求cos2Xo的值
已知函数f(x)=2√ 3sinxcosx+2cos^2x-1若f(Xo)=6/5,Xo属于【π/4,π/2】,求cos2Xo的值
f(x)=√3sin2x+cos2x=2sin(2x+π/6) sin(2x0+π/6)=3/5
Xo属于【π/4,π/2】2x0+π/6∈[2π/3,7π/6] cos(2x0+π/6)=-4/5
cos2Xo=cos[(2x0+π/6)-π/6]=-4√3/10 +3/10=(3-4√3)/10
2cos^2x是2(cosx)^2么?
f(x)=2√ 3sinxcosx+2cos^2x-1
=√ 3sin2x+cos2x
=2sin(2x+π/6)
wo 知道啊,哥们,你先给分,我马上传给你,谢
f(x)=2√ 3sinxcosx+2cos^2x-1=√ 3sin2x+cos2x=2sin(2x+π/6)
由Xo属于【π/4,π/2】得2π/3=<2x+π/6=<7π/6
故由f(Xo)=6/5得sin(2x0+π/6)=5/12,则cos(2Xo+π/6)=-√119/12
则
cos2Xo=cos(2Xo+π/6—π/6)
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f(x)=2√ 3sinxcosx+2cos^2x-1=√ 3sin2x+cos2x=2sin(2x+π/6)
由Xo属于【π/4,π/2】得2π/3=<2x+π/6=<7π/6
故由f(Xo)=6/5得sin(2x0+π/6)=5/12,则cos(2Xo+π/6)=-√119/12
则
cos2Xo=cos(2Xo+π/6—π/6)
=cos(2Xo+π/6)cosπ/6+sin(2x0+π/6)sinπ/6
=(5-√357)/24
输入的太辛苦了
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