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求下列三角函数值(1)cos-1109°(2)sin-1050°(3)tan19π/3
人气:226 ℃ 时间:2020-05-12 02:43:17
解答
(1) cos(-1109°=cos1109 °.=cos(3*360 °+29°).=cos29°.( =0.8746).(2)sin(-1050)=-sin1050.=-sin(1080°-30°).=-sin(3*360°-30).=-(-sin30°.=sin30.=1/2.( =0.50°).(3) tan(19π/3)=tan(6π+π/3).=tan(3...
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