对y=sinx求导 y`=cosx 将x=4/Π代入 y`=cos(4/Π) 设切线方程f(x)=cos(4/Π)x+b1 代入点[4/Π,sin(4/Π)] 所以b1=sin(4/Π)-(4/Π)cos(4/Π) 所以切线方程为 f(x)=cos(4/Π)x+sin(4/Π)-(4/Π)cos(4/Π) 设法线方程...