求曲线y=sinx 在x=4/Π 处的切线方程和法线方程；拜托了各位

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) 设法线方程为g(x)=1/[cos(4/Π)]x+b2 代入点[4/Π,sin(4/Π)] 所以b2=sin(4/Π)-(4/Π)/cos(4/Π) 所以法线方程为g(x)=1/[cos(4/Π)]x+sin(4/Π)-(4/Π)/cos(4/Π) QQ849123250