用公式一求下列角的三个三角函数值：21π/4,-23π/6,
<p>问题：用公式一求下列角的三个三角函数值：21π/4,-23π/6,<p>答案：↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">卢结成的回答：<div class="content-b">网友采纳 sin(21π/4)=sin(4π+5π/4)=sin(π+π/4)=-sin(π/4)=-√2/2 sin(-23π/6)=sin(-4π+π/6)=sin(π/6)=0.5 cos(21π/4)=cos(4π+5π/4)=cos(π+π/4)=-cos(π/4)=-√2/2 cos(-23π/6)=cos(-4π+π/6)=cos(π/6)=√3/2 tan(21π/4)=tan(4π+5π/4)=tan(π+π/4)=tan(π/4)=1 tan(-23π/6)=tan(-4π+π/6)=tan(π/6)=√3/3
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