fyqt.net
当前位置:首页 >> 1 Cos50 tAn10 >>

1 Cos50 tAn10

=1/cos50°+sin10°/cos10° =1/sin40°+cos80°/sin80° =2cos40°/(2sin40°cos40°)+cos80°/sin80° =(2cos40°+cos80°)/sin80° =【2cos(10°+30°)+cos80°】/sin80° =(2cos10°cos30°-2sin10sin30°+cos80°)/sin80° =(2cos30°cos10°-sin10°+cos80°)/sin...

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

(cos40°+sin50°(1+√3tan10°))÷(sin70°√(1+cos50°)) =(cos40°+sin50°(1+√3sin10°/cos10°))÷(sin70°√(2(cos25°)^2) =(cos40°+sin50°(cos10°+√3sin10°)/cos10°)÷(√2sin70°cos25°) =(cos40°+2sin50°sin40°/cos10°)÷(√2sin70°cos25...

原式=2sin50°+cos10°+3sin10°cos35°cos40°+sin35°sin40°=2sin50°+2cos50°cos5°=22(22sin50°+22cos50°)cos5°=22sin95°cos5°=22cos5°cos5°=22.

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...

【1】单位:度“º”。【2】tan50=tan(60-10)=(tan60-tan10)/(1+tan60tan10)=(tan60-tan10)/[1+(√3)tan10].===>1+(√3)tan10=(tan60-tan10)/tan50=(tan60-tan10)cos50/sin50=(tan60-tan10)cos50/cos40.===>[1+(√3)tan10]cos40=(tan60-tan10)cos...

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°+...

(sin50-cos50tan10)cos40 =sin50*cos40-cos50*cos40*sin10/cos10 =cos40*cos40-sin40*cos40*sin10/sin80 =(cos40)^2-1/2 (sin80)*sin10/sin80 =1/2(1+cos80)-1/2(sin10) =1/2+1/2(sin10)-1/2(sin10) =1/2

显然1+根号3tan10 =1+tan60tan10 =(cos60cos10+sin60sin10)/cos60cos10 =cos(60-10)/cos60cos10 =cos50/cos60cos10 =2cos50/cos10 而1+cos20=2(cos10)^2 故得到根号(1+cos20)=根号2 *cos10 原式=(2sin50+ 2sin10*cos50/cos10) * (根号2 *cos10) ...

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