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sin(-1200°).cos1209°+cos(-1020°)-sin(-1050°)+tan855°
人气:262 ℃ 时间:2020-06-19 19:17:22
解答
sin(-1200)
=sin(-4*360+240)
=sin240
=sin(180+60)
=-sin60
=-√3/2
cos1209+cos(-1020)-sin(-1050)+tan855°
=cos(360*3+129)+cos1020+sin1050+tan(4*180+135)
=cos129+cos(3*360-60)+sin(3*360-30)+tan135
=cos129+cos60-sin30+tan135
=cos129+1/2-1/2-1
=cos129-1
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