fyqt.net
当前位置:首页 >> 1 Cos50 tAn10 >>

1 Cos50 tAn10

1/cos50°+tg10°=1/sin40° + sin10°/cos10° =1/sin40° +(sin10°*2sin10°) /(cos10°*2sin10°) =1/sin40° +2(sin10°)^2 /sin20° =1/sin40° +(1-cos20°) /sin20° =1/sin40°+(1-cos20°)*2cos20° /sin20°*2cos20° =[1+2cos20°-2(cos20°)^2] /sin40° =(...

解: 1/(cos50°)+tan10°-2√3 ≈1.55572382686+0.176326981-2x1.732051 =-1.73205081 ≈-√3

1

原式=(cos10°+√3sin10°)/cos 50° =2sin(10°+30°)/cos 50° =2sin40°/cos 50° =2

因为 tan50°=tan(60°-10°) =(tan60°-tan10°)/(1+√3tan10°) sin50*(1+tan60*tan10) =cos50°*(tan60°-tan10°)*(1+√3*tan10)/(1+√3tan10°) =cos50°(tan60°-tan10°) =cos50°(sin60°/cos60°-sin10°/cos10°) =cos50° (sin60°cos10°-sin10°cos60°)/(cos...

原式=(2sin50º+cos10º+√3sin10º)/√2cos5º =[2sin50º+2(1/2*cos10º+√3/2sin10º)]/√2cos5º =(2sin50º+2cos50º)/√2cos5º=2√2(√2/2sin50º+√2/2cos50º)/√2cos5º=2√2sin95&o...

(sin80°/cos50°)(1+√3tan10°) =(cos10°/cos50°)(1+√3*sin10°/cos10°) =(cos10°+√3*sin10°)/cos50° =2[(1/2)cos10°+(√3/2)*sin10°]/cos50° =2(cos60°*cos10°+sin60°*sin10°)/cos50° =2cos(60°-10°)/cos50° =2

cos40°+sin50°×(1+√3tan10°) =cos40°+sin50°×(tan60°-tan10°)/tan50° =cos40°+(tan60°-tan10°)cos50° =cos40°+√3cos50°-tan10°cos50° =cos40°+√3sin40°-tan10°sin40° =2[(1/2)cos40+(√3/2)sin40°]-(sin10°/cos10°)sin40° =2(cos60°cos40°+...

=cos10(tan10-√3)/sin(60-10) =sin10-√3cos10/(√3cos10-sin10/2) =-2

网站首页 | 网站地图
All rights reserved Powered by www.fyqt.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com