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求下列三角函数的值(1)cos(-1050°)(2)tan19/3π
人气:168 ℃ 时间:2020-10-01 18:55:34
解答
(1)cos(-1050°)
=cos(-1050°+3*360°)
=cos(-1050°+1080°)
=cos30°
=√3/2
(2)tan19/3π
=tan(6π+π/3)
=tanπ/3
=√3
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