求下列各三角函数值tan(-17π/6),tan(-31π/4)

学习 时间:2026-08-16 07:19:34 阅读:5486
求下列各三角函数值tan(-17π/6),tan(-31π/4)

最佳回答

标致的嚓茶

无语的溪流

2026-08-16 07:19:34

-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

最新回答共有2条回答

  • 洁净的香氛
    回复
    2026-08-16 07:19:34

    -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

上一篇 No wonder that you are scolded;you were out enjoying yoursel

下一篇 求科普 这是什么东西 干什么使的.