Question

求曲线y=sin(π/2-x),在点A(-π/3,1/2)处的切线方程,求详解,..

Answer 1

y=sin(π/2-x)=cosx

y'=-sinx,y'(-π/3)=-sin(-π/3)=√3/2.

所以,所求切线方程为：y-1/2=(√3/2)(x+π/3),即√3x-2y+1+√3π/3=0.