sin(-1200°)-cos1290°+cos(-120)°+cos(-1020)°-sin(-1050°)+tan855°怎么做？最好有过程
答案:2 悬赏:80
解决时间 2021-11-08 19:34
- 提问者网友：回憶丶初
- 2021-11-08 15:23
sin(-1200°)-cos1290°+cos(-120)°+cos(-1020)°-sin(-1050°)+tan855°怎么做？最好有过程
最佳答案
- 二级知识专家网友：woshuo
- 2021-11-08 16:53
sin(-1200°)=-sin(3*360° +120° )=-sin120°=-1/2
cos1290°=cos(4*360°-150°)=cos150°=-√3/2
cos(-120)°=cos120°=-1/2
cos(-1020)°=cos1020°=cos(3*360°-60°）=1/2
sin(-1050°)=sin(3*360°-30°）=-sin30°=-1/2
tan855°=tan135°=-1
sin(-1200°)-cos1290°+cos(-120)°+cos(-1020)°-sin(-1050°)+tan855°
=-1/2-（-√3/2）-1/2+1/2-（-1/2）+（-1）=（√3-2）/2
cos1290°=cos(4*360°-150°)=cos150°=-√3/2
cos(-120)°=cos120°=-1/2
cos(-1020)°=cos1020°=cos(3*360°-60°）=1/2
sin(-1050°)=sin(3*360°-30°）=-sin30°=-1/2
tan855°=tan135°=-1
sin(-1200°)-cos1290°+cos(-120)°+cos(-1020)°-sin(-1050°)+tan855°
=-1/2-（-√3/2）-1/2+1/2-（-1/2）+（-1）=（√3-2）/2
全部回答
- 1楼网友：伤口狠精致
- 2021-11-08 17:10
sin(-1200°)cos1290°+cos(-1020°）sin（-1050°)+tan855
=sin240°cos210°+cos60°sin30°+tan135°
=1-1=0
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