y=sinx y'=cosx 所以在点(π/6,1/2)处的切线斜率是k=cos(π/6)=√3/2 所以在点(π/6,1/2)处的法线斜率是k=-1/(√3/2)=-2√3/3 所以切线方程是y-1/2=(√3/2)*(x-π/6)即y=(√3/2)*(x-π/6)+1/2 法线方程是y-1/2=(-2√3/3)*(x-π/6) ,6,
