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化简变形f(x)=[2sin(x+π/3)+sinx]cosx-√3sin^2x
人气:270 ℃ 时间:2020-01-31 10:27:03
解答
=[2 sinx cosπ/3 +2cosx sinπ/3+sinx ]cosx-√3sin^2x=[sinx+√3cosx+sinx]cosx-√3sin^2 x=2sinxcosx+√3(cos^2x-sin^2x)=sin 2x +√3cos 2x=2(sin2x cosπ/3+ cos 2x sin π/3)=2sin(2x+π/3)
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