llgd.net
当前位置:首页 >> tAn15度是多少?? >>

tAn15度是多少??

tan15° =tan(60°-45°) =(tan60°- tan45°)/(1+ tan60°*tan45°) =(根号3 -1)/(1+根号3) =(根号3-1)*(根号3-1)/[(根号3+1)(根号3-1)] =(4-2根号3)/2 =2-根号3 亲,答题不易。记得好评奥~

tan15° =tan(60°-45°) =(tan60°-tan45°)/(1+tan60°tan45°) =(√3-1)/(1+√3) =(√3-1)²/[(1+√3)(√3-1)] =(4-2√3)/2 =2-√3

求采纳

答: tan15° =tan(60°-45°) =(tan60°- tan45°)/(1+ tan60°*tan45°) =(根号3 -1)/(1+根号3) =(根号3-1)*(根号3-1)/[(根号3+1)(根号3-1)] =(4-2根号3)/2 =2-根号3

tan15° =tan(60°-45°) =(tan60°- tan45°)/(1+ tan60°*tan45°) =(根号3 -1)/(1+根号3) =(根号3-1)*(根号3-1)/[(根号3+1)(根号3-1)] =(4-2根号3)/2 =2-根号3 tan60°=根号3 tan60°*tan15°=根号3*(2-根号3)=2*根号3-3

最简单的是

几何方法 先做一个含有30°的直角三角形,∠C=90°,∠ABC=30° 延长CB到点D,使BA=BD ∵∠ABC=30° ∴∠BAD+∠D=30° ∵AB=BD ∴∠BAD=∠D=15° 设AC=X,则BD=AB=2x BC=根号3*x tan∠D=X/(2+根号3)x=2-根号3 也就是 tan15°=2-根号3 而根号3≈1.732, ∴tan∠D≈0.268

tan15° =tan(60°-45°) =(tan60°- tan45°)/(1+ tan60°*tan45°) =(√3 -1)/(1+√3) =(√3-1)*(√3-1)/[(√3+1)(√3-1)] =(4-2√3)/2 =2-√3

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