Maximun, minimum and inflection points of a function

We will solve the problem without having the graph of the sine.

Maxima / minimums.

First of all, we compute the derivative of the sine and we find its roots: $$y'=cos(x)=0 \Rightarrow x=\pm\dfrac{\pi}{2},\pm\dfrac{3\pi}{2},\pm\dfrac{5\pi}{2},\pm\dfrac{7\pi}{2},\ldots$$

We compute the sign of each of the solutions to determine if it is a maximum or a minimum: $y''=-sin(x)$. $$x=\dfrac{\pi}{2} \rightarrow y''(\dfrac{\pi}{2})=-1 < 0 \Rightarrow Max$$ $$x=\dfrac{3\pi}{2} \rightarrow y''(\dfrac{3\pi}{2})=1 > 0 \Rightarrow Min$$ $$x=\dfrac{5\pi}{2} \rightarrow y''(\dfrac{5\pi}{2})=-1 < 0 \Rightarrow Max$$ $$\ldots$$

The values of the function in the maximum is $1$ and in the minimum is $-1$.

Inflection points.

The second derivative is equal to zero: $$y''=-sin(x)=0 \Rightarrow x=0,\pm\pi,\pm2\pi,\pm3\pi,\ldots$$ $$y(x)=sin(x)=0$$

See now the graph: