Question

How do you find the equation of the line tangent to `y = 4 cos x` at x = π/6?

Answer 1

The equation of the tangent at `x=pi/6` for `y=4*cosx` is

`y-y(pi/6)=y'(pi/6)(x-pi/6)`

where `y(pi/6)=4*cos(pi/6)=4*sqrt3/2=2*sqrt3`

and `y'(x)=-4*sinx` and `y'(pi/6)=-4sin(pi/6)=-4*1/2=-2`

Hence we have that

#y-2*sqrt3=-2*(x-pi/6)=> y=-2x+pi/3+2*sqrt3#