关于三角函数...此题的简便算法是什么？请教数学帝

显示全部楼层 · 2010-8-22 08:18:12

cos2t = 2cost^2 - 1 = 1 - 2sint^2所以：分子
= 2(sinx/2)^2 + 2 sin(x/2) * cos(x/2) = 2sin(x/2)*(sin(x/2)+cos(x/2))同理：分母= 2cos(x/2)*(sin(x/2)+cos(x/2))所以 tan(x/2) = 2;tanx = 2tan(x/2)/(1-tan(x/2)^2) = -4/3

千问 · 2010-8-22 08:18:12

解：（1-cosx+sinx）/（1+cosx+sinx）=21-cosx+sinx=2+2cosx+2sinx3cosx+sinx+1=0两边除以cosx，得3+tanx+1/cosx=0，即3+tanx+secx=01+tanxtanx=secxsecx1+tanxtanx=〔-（3+tanx）〕^2