Question

find the equation of the tangent lines at the zeros of f(x)=sinx on the interval [0,2pi]

Answer 1

f ' (x) = cos x

at x = 0, cos x = 1

Equation of tangent line is y - 2pi = 1(x - 0)

or, y - 2 pi = x

or y = x + 2pi

Keep in mind, that the derivative is the slope of the tangent line. So cos x is the slope of sin x. At x = 0, the slope is 1.