设函数f(x)sin(x+π/3)+2sin(x+π/3)-根号3cos(2π/3-x) （1）求f(π/6),f(π/3)的值由（1）你能得到什么结论?

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设函数f(x)sin(x+π/3)+2sin(x+π/3)-根号3cos(2π/3-x) （1）求f(π/6),f(π/3)的值由（1）你能得到什么结论?
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设函数f(x)sin(x+π/3)+2sin(x+π/3)-根号3cos(2π/3-x) （1）求f(π/6),f(π/3)的值由（1）你能得到什么结论?
设函数f(x)sin(x+π/3)+2sin(x+π/3)-根号3cos(2π/3-x) （1）求f(π/6),f(π/3)的值
由（1）你能得到什么结论?

设函数f(x)sin(x+π/3)+2sin(x+π/3)-根号3cos(2π/3-x) （1）求f(π/6),f(π/3)的值由（1）你能得到什么结论?
题目是 f(x=sin(x+π/3)+2sin(x+π/3)-√3cos(2π/3-x) !

（1）f（π/6）=0；f(π/3）=0 【直接代入，很容易得出】
（2）f（x）=0
∵2sin(x-π/3) = 2sin[(x+π/3)-2π/3]
= 2sin(x+π/3)cos(2π/3)-2cos(x+π/3)sin(2π/3)
= -sin(x+π/3)-√3cos(x+π/3)
且cos(2π/3-x) = -cos[π-(2π/3...

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（1）f（π/6）=0；f(π/3）=0 【直接代入，很容易得出】
（2）f（x）=0
∵2sin(x-π/3) = 2sin[(x+π/3)-2π/3]
= 2sin(x+π/3)cos(2π/3)-2cos(x+π/3)sin(2π/3)
= -sin(x+π/3)-√3cos(x+π/3)
且cos(2π/3-x) = -cos[π-(2π/3-x)] = -cos(x+π/3)
∴有sin(x+π/3)+2sin(x-π/3)- √3cos(2π/3-x)
= sin(x+π/3)+[-sin(x+π/3)-√3cos(x+π/3)]-√3[-cos(x+π/3)] 【将上述代入】
= 0

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