求下列各三角函数值tan(-17π/6),tan(-31π/4)

学习 时间:2026-06-04 18:19:53 阅读:5131
求下列各三角函数值tan(-17π/6),tan(-31π/4)

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2026-06-04 18:19:53

-17π/6=-18π/6 + π/6=-3π +π/6因此:tan(-17π/6)=tan(-3π +π/6)=tan(-π+π/6)=tan(-5π/6)=-tan(5π/6)=-tan(π - π/6)=tan(π/6)=√3/3-31π/4=-32π/4+π/4=-8π+π/4因此:tan(-8π+π/4)=tanπ/4=1

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  • 动听的啤酒
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    2026-06-04 18:19:53

    -17π/6=-18π/6 + π/6=-3π +π/6因此:tan(-17π/6)=tan(-3π +π/6)=tan(-π+π/6)=tan(-5π/6)=-tan(5π/6)=-tan(π - π/6)=tan(π/6)=√3/3-31π/4=-32π/4+π/4=-8π+π/4因此:tan(-8π+π/4)=tanπ/4=1

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