x=tan(t) sin(t)-cos(t)=?

x=tan(t) sin(t)-cos(t)=? cosθ=t,求sinθ,tanθ 知道cos t,怎么求tan cos t=(1-x^2)^(1/2)tan t=(x)/((1-x^2)^(1/2))我想知道那个tan t是怎么求出来的t=arc(sin x)用tan t^2=1/(1+cos t^2) 公式好像算不出来啊，我算出来是(tan t)^2=1/[1+(1-x^2)] 设cosθ=t(t≥0),求sinθ和tanθ的值 已知tanα=t 用t表示sinαcosα, 设tan(θ/2)=t,求证：sinθ=2t/(1+t^2),cosθ=(1-t^2)/(1+t^2),tanθ=2t/(11t^2)设tan(θ/2)=t,求证：sinθ=2t/(1+t^2),cosθ=(1-t^2)/(1+t^2),tanθ=2t/(1-t^2) matlab中怎么对下面的程序进行二维插值并画出三维视图呢?[x,y]=meshgrid(-2:1:2);t=2;Z=[(sin(t))^3 t^2 sin(t) cos(t) (sin(t))^2;t^2 sin(t) (cos(t))^4 cos(t) (sin(t))^2;t sin(t) cos(t) t^2 (sin(t))^2;sin(t) (cos(t))^3 t^2 (sin(t))^2 (sin t) / ( 1-cos t)= 已知tanα=t 求 sin^2α-4sinαcosα+3cos^2α) sin (ωt) / cos (ωt) = sin (ωt) / sin(ωt + π/2) = e ^ (-j.π/2) = -j 对么?sin (ωt) / cos (ωt) = sin (ωt) / sin(ωt + π/2) = e -j.π/2 = cos(-π/2) + j.sin (-π/2) = -j ,tan (ωt) = -j 那么 tan (ωt) = -j 但是sin (ωt) = (e j.ωt - e -j. 急证：sin t + cos t - 2/[1+tan^2 (t/2) ] tan t/2怎么算出等于sin t /（1+cos t）. 已知tanα=t,求sinα,cosα的值 为什么cos(sin(tan x)))=1 matlab微分方程画图问题已知：x'[t] = 2 (Sin[t] - x[t]) /Sqrt[(Sin[t] - x[t])^2 + (Cos[t] - y[t])^2]y'[t] = 2 (Cos[t] - y[t]) /Sqrt[(Sin[t] - x[t])^2 + (Cos[t] - y[t])^2]初始x[0]=0y[0]=0画出（x,y）的轨迹 设tan(θ/2)=t,求证：sinθ=2t/(1+t^2),cosθ=(1-t^2)/(1+t^2),tanθ=2t/(1-t^2)sec没有学过 设tan(θ/2)=t,求证：sinθ=2t/(1+t^2),cosθ=(1-t^2)/(1+t^2),tanθ=2t/(1-t^2) MATLAB 向量相乘的问题已知 t=0:1:20;A=[1,0,0;0,1,0;0,0,1];B=[-sin(t),cos(t),sin(t);-t,t,tan(t);sin(t),cos(t),cos(t)];C=[1,0,1;0,1,0;0,0,1];如何求A*B*C