y=sin^2(x)+2sin(x)cos(x)+3cos^2(x)的最值 用向量解
y=sin^2(x)+2sin(x)cos(x)+3cos^2(x)的最值 用向量解
sin^2x+cos^2y=1/2 求3sin^2x+sin^2y的最值
化简y=sin^2(x)+2sin(x)cos(x)+3cos^2(x)
已知向量a=(cosωx-sinωx,sinωx),b=(-cosωx-sinωx,2√ 3cosωx),
设sin(x+y)sin(x-y)=m,则cos^2x-cos^2y的值
Sin x-sin y=2/3 cos x-cos y=1/2 求cos(x-y)
y =(cos^2) x - sin (3^x),求y'
sin(x+y)sin(x-y)=k,求cos^2x-cos^2y
求导f(x) = cos(3x) * cos(2x) + sin(3x) * sin(2x).
证明COS(X+Y)COS(X-Y)=COS^2X-SIN^2Y
已知函数y=(sin x+ cos x)(sin x+cos x)+2cos x*cos x ,求它的递减区间
已知tan=2,求(cos x+sin x)/(cos x-sin x)+sin^2x