求下列三角函数值(1)cos-1109°(2)sin-1050°

求下列三角函数值(1)cos-1109°(2)sin-1050°（3）tan19π/3

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(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.
（亩禅悉 =0.50°）迅乎袭大.
(3) tan(19π/3)=tan(6π+π/3).
=tan(3*2π+π/3).
=tanπ/3.
=√3.
( =1.732).

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(1)cos(-1109˚)=cos1109˚=cos(360˚*3+29˚)=cos29˚(不是特殊灶乱角，请自己前握按计算?)
(2)sin(-1050˚)=-sin1050˚=-sin(1080˚-30˚慧辩庆)=-sin(-30˚)=sin30˚=1/2
(3)tan19π/3=tan(6π+π/3)=tanπ/3=√3

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