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I was just messing around is Desmos and plotted the graph of $$y=\frac{\frac{\tan(x)}{\sin(x)}}{\cos(x)}$$ and I realised that all the minimum points on the graph were multiples of $\pi$. Could someone please explain how this is so?

enter image description here

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  • $\begingroup$ Have you tried simplifying your function? And do you know derivatives? $\endgroup$ – Randall Mar 30 '18 at 14:48
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    $\begingroup$ Rearranging, $y = \sec^2(x)$, which has a period of π. That is, if you use radians, and not degrees. $\endgroup$ – Kaynex Mar 30 '18 at 14:49
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When you are pfaffing about with mathematical ideas, you often come up with things that look mysterious and strange. This is good! This means that there is something to learn! However, humans are not very good at attacking difficult ideas directly, so we have to try to make the problem more simple to understand. Looking at that fraction, I immediately see two things that make me uncomfortable: (1) there are a ton of trigonometric functions running around and (2) that fraction is gross. Let's start by trying to simplify this:

\begin{align} \frac{\frac{\tan(x)}{\sin(x)}}{\cos(x)} &= \tan(x) \cdot \frac{1}{\sin(x)} \cdot \frac{1}{\cos(x)} && \left( \text{properties of fractions} \right) \\ &= \frac{\sin(x)}{\cos(x)} \cdot \frac{1}{\sin(x)} \cdot \frac{1}{\cos(x)} && \left( \text{definition of $\tan$} \right) \\ &= \frac{1}{\cos(x)} \cdot \frac{1}{\cos(x)} && \left( \text{cancel the $\sin$ terms} \right) \\ &= \frac{1}{\cos(x)^2}. \end{align} As has been noted by others, this could be written as $\sec(x)^2$, but I prefer to use $\frac{1}{\cos(x)^2}$, since the secant function is a bit mysterious to me, and I am more comfortable working with sines and cosines when possible. This is, of course, a matter of taste.

Now, instead of wondering why the original crazy fraction has minima at integer multiples of $\pi$, we can ask why the function $$ x \mapsto \frac{1}{\cos(x)^2} $$ has minima at integer multiples of $\pi$. There is an argument which can be made using the tools of calculus, but it might be more appropriate to make a more basic geometric argument. Notice that since the numerator of the fraction is $1$ (i.e. a constant), the fraction is minimized when the denominator is maximized. That is, we are interested in the values of $x$ such that $\cos(x)^2$ is maximal.

But $\cos(x)^2$ is maximized when $|\cos(x)|$ is maximized—that is, when $\cos(x)$ attains its largest positive and largest negative values. From the unit circle definition of the cosine function, we know that (a) the cosine of an angle (measured in radians) represents the $x$-coordinate of a point on the unit circle corresponding to that angle, and (b) the largest possible magnitude of an $x$-coordinate of a point on the unit circle is $1$; this happens at the angles $0$ and $\pi$ (plus any number of additional rotations). This means that $\cos(x)$ is maximal for any number of the form $$ 0 + 2k\pi \qquad\text{or}\qquad \pi + 2k\pi, $$ where $k$ is any integer. But we have $$ \{ 0 + 2k\pi : k \in \mathbb{Z} \} \cup \{\pi + 2k\pi : k\in \mathbb{Z} \} = \{ n\pi : n\in\mathbb{Z} \}, $$ therefore the cosine function is maximized at integer multiples of $\pi$.

Putting all of this back together, we conclude that the minimum values of $$ y = \frac{\frac{\tan(x)}{\sin(x)}}{\cos(x)} $$ occur when $x$ is an integer multiple of $\pi$. Moreover, since $\cos(x)^2 = 1$ for such values of $x$, the minimum value of $y$ is $1$.

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It is $1/\cos^2(x)$ and $|\cos(x)|$ has its maximal values in $k\pi$.

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