已知f(x)=sin(π-x)cos(2π-x)tan( -x+3/2π)除以cos( - π- x),则f(- 31/3 π)的值?
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![已知f(x)=sin(π-x)cos(2π-x)tan( -x+3/2π)除以cos( - π- x),则f(- 31/3 π)的值?](/uploads/image/z/2496817-1-7.jpg?t=%E5%B7%B2%E7%9F%A5f%EF%BC%88x%EF%BC%89%3Dsin%EF%BC%88%CF%80-x%EF%BC%89cos%EF%BC%882%CF%80-x%EF%BC%89tan%EF%BC%88+-x%2B3%2F2%CF%80%EF%BC%89%E9%99%A4%E4%BB%A5cos%EF%BC%88+-+%CF%80-+x%EF%BC%89%2C%E5%88%99f%EF%BC%88-+31%2F3+%CF%80%EF%BC%89%E7%9A%84%E5%80%BC%3F)
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已知f(x)=sin(π-x)cos(2π-x)tan( -x+3/2π)除以cos( - π- x),则f(- 31/3 π)的值?
已知f(x)=sin(π-x)cos(2π-x)tan( -x+3/2π)除以cos( - π- x),则
f(- 31/3 π)的值?
已知f(x)=sin(π-x)cos(2π-x)tan( -x+3/2π)除以cos( - π- x),则f(- 31/3 π)的值?
sin(π-x)=sinx,
cos(2π-x)=cosx,
tan( -x+3/2π)=tan(π/2-x)=cotx=cosx/sinx,
cos( - π- x)=cos(π-x)=-cosx,
则f(x)=cos²x/-cosx=-cosx,
f(- 31/3 π)=-cos(-31/3π)=-cos(-π/3)=-cosπ/3=-1/2.
我会 ,就不告诉你