show that $y = \cos x$, has a maximum turning point at $(0, 1)$ and a minimum turning point at $(\pi, -1)$ 
show that  $y = \cos x$, has a maximum turning point at $(0, 1)$ and a
  minimum turning point at $(\pi, -1)$

Turning points occur when the gradient is 0 or $\frac{dy}{dx} = 0$
$f(x) = \cos x$
$\frac{d}{dx} = - \sin x$
$x = -\sin^{-1}(0)$
$x = 0$
sin is in the 1st and 2nd quadrant so $x = 0$ or $x = -90$
$f''(x) = - \cos x$
$f''(0) = - 1$ which is a maximum turning point
$f''(-90) = 0$, is this a point of inflection so I am not sure if this is right.
When $x = 0, f(x) = \cos(0) = 1$
So we have a maximum turning point at $(0, 1)$.
I'm not sure how to get the minimum turning point?
 A: 
$$f(x)=\cos x \qquad [-1,\pi]$$

Finding the $\min$ and $\max$:
$$f'(x)=-\sin x$$
that's equal to zero when $x=\color{red}{0},x=\color{red}\pi$ (critical points)
$$f''(x)=-\cos x\\
f''(0)=-\cos 0=-1$$
$\Longrightarrow\quad x=0$ is $\max$ of $\cos x$ 
$$\boxed{\max \text{ point }(0,1)}$$
$$f''(\pi)=1$$
$\Longrightarrow\quad x=\pi$ is $\min$ of $\cos x$ 
$$\boxed{\min \text{ point }(\pi,-1)}$$
A: The derivative of $f(x) = \cos x$ is $f'(x) = -\sin x$.  The critical points of $f$ occur where $f'(x) = 0 \implies -\sin x = 0 \implies \sin x = 0$.  
The sine of an angle $\theta$ in standard position (vertex at the origin, initial side on the positive $x$-axis) is the $y$-coordinate of the point where the terminal side of the angle intersects the unit circle.

Consequently, $\sin\theta = 0$ if the terminal side of angle $\theta$ lies on the $x$-axis.  
Hence, $\sin x = 0 \implies x = n\pi$, where $n$ is an integer.  Thus, $x = 0$ and $x = \pi$ are critical points of the function $f(x) = \cos x$.
We can draw the sine graph by tracing out the $y$-coordinates of the point where the terminal side of the angle intersects the unit circle.
 
The graph of $f'(x) = -\sin x$ is obtained by reflecting the graph of the sine function in the $x$-axis.

From the graph of $f'(x) = -\sin x$, we see that the derivative is positive in the interval $[-\pi/2, 0)$ and negative in the interval $(0, \pi/2]$.  Thus, the function $f(x) = \cos x$ is increasing to the left of $0$ and decreasing to its right.  Hence, $f(x) = \cos x$ has a relative maximum at $x = 0$ by the First Derivative Test. The relative maximum value is 
$f(0) = \cos 0 = 1$.  
From the graph of $f'(x) = -\sin x$, we see that the derivative is negative in the interval $[\pi/2, \pi)$ and positive in the interval $(\pi, 3\pi/2]$.  Thus, the function $f(x) = \cos x$ is decreasing to the left of $\pi$ and increasing to its right.  Hence, $f(x) = \cos x$ has a relative minimum at $x = \pi$ by the First Derivative Test.  The relative minimum value is $f(\pi) = \cos \pi = -1$, as shown in the graph below.

