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f(x)=[2sin(x+π/3)+sinx]cosx-根号3sin2x,化简
人气:192 ℃ 时间:2020-06-07 22:57:35
解答
f(x)=[2sin(x+π/3)+sinx]cosx-√3sin2x
=[2sinx+√3cosx]cosx-√3sin2x
=sin2x+(√3/2)(1+cos2x)-√3sin2x
=(1-√3)sin2x+(√3/2)cos2x+√3/2
=繁!
请检查题目f(x)=[2sin(x+π/3)+sinx]cosx-根号3sin^2x
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