Maximum and minimum of $f(x)= \sin x(1- \cos x)$ at $[0,2\pi] \to [-2,2|$

Question

Let $f:[0,2\pi] \to [-2,2|$ with

$$f(x)= \sin x(1- \cos x)$$

Using WolframAlpha I checked that the maximum is at $2\pi n + (2\pi/3)$ and the minimum at $2\pi n - (2\pi/3)$

But how can I get to that by calculation?

Answer 1

Take derivative of $f(x)= \sin x(1- \cos x)$ and set it to zero,

$$\cos x (1-\cos x)+\sin^2 x=0$$

or,

$$2\cos^2x-\cos x-1=0$$

which has the solutions,

$$\cos x = -\frac12, \>\>\>\>\> \cos x =1$$

Answer 2

Differentiate and find solutions for $f’(x)=0$, then conclude by second derivative test https://en.m.wikipedia.org/wiki/Derivative_test

$f’’(x)<0$ then max

$f’’(x)>0$ then min

$f’’(x)=0$ then you can’t tell

Answer 3

Hint:

Factor the derivative: $$f'(x)=\cos x(1-\cos x)+\sin^2x=\cos x -\bigl(\cos^2x-\sin^2x\bigr)=\cos x-\cos 2x=2\sin\frac{3x}2\sin\frac x2$$ and study its sign on $[0,2\pi]$.