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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?

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    $\begingroup$ Is the value of $ \ x \ $ given by the inverse sine function really the only value for which $ \ \frac{dy}{dx} \ = \ 0 \ $ ? (Also, $ \ -90º \ $ is neither in the second quadrant, nor is its sine value zero.) $\endgroup$ – colormegone Apr 16 '16 at 6:34
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$$f(x)=\cos x \qquad [-1,\pi]$$

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 $\cos x$

$$\boxed{\min \text{ point }(\pi,-1)}$$

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  • $\begingroup$ $\sin (0) = \pi$. how is that? $\endgroup$ – dagda1 Apr 16 '16 at 7:33
  • $\begingroup$ @dagda1 I didn't say that, I said $-\sin 0=0$ and also $-\sin \pi=0$ $\endgroup$ – 3SAT Apr 16 '16 at 8:54
  • $\begingroup$ $- \sin \pi = 0$?. This because sin is positive in the 1st and 2nd quarter and 180 - 0 = 180 or \pi? $\endgroup$ – dagda1 Apr 16 '16 at 11:03
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    $\begingroup$ @dagda1 do you agree that $\sin \pi=0$? multiply this by $-1$, still zero $\endgroup$ – 3SAT Apr 16 '16 at 12:42
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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.

circular_function_definitions

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.

graph_of_sine_function

The graph of $f'(x) = -\sin x$ is obtained by reflecting the graph of the sine function in the $x$-axis.

graph_of_negative_sign_function

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.

graph_of_cosine_function

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