On The Shape Of Trig Graphs To A Precalc Student I want to know why 3 types of trig graphs have the shape that they do: 


*

*Secant graphs. Why would $\dfrac {1}{\cos \theta}$ result in a graph like this? I get the cosine graph - it makes sense when you compare it to a unit circle. But I don't understand it at the same level for secant graphs. Can someone please explain why the secant graph is shaped like that given it's equation? 

*Cosecant graphs. Same as above.

*Cotangent graphs. Why would $\dfrac {1}{\tan \theta}$ cause the tangent graph to flip? Because when you look at a cotangent graph, it's basically a reflected tangent graph. Why would $\dfrac {1}{\cos \theta}$ create a graph that looks like that? 
Can you please give the explanation for why the graphs are shaped like that not too rigorously, and at the level of a Precalculus student who hasn't learnt Calculus yet? Thank you.
 A: consider how behavior in a graph of $f(x)$ will affect behavior in a graph $\frac 1 {f(x)}$.  For values where $0 < f(x) < 1$ then we will have $\infty > \frac 1{f(x)}$, with little changes from $teensy$ to $\frac 12 teensy$ resulting in huge stretches from $huge$ to $2 \times huge$. And if $f(x)$ goes through from $teensy$ to $0$ to $-teensy$ then $\frac 1{f(x)}$ will get asymptotic to infinity, be infinite(undefined) at a point, and jump to "negative infinity" and then imediate start reducing to more moderate negative values as $f(x)$ because more siginificant negative values.
If $1 < f(x)$ we will have $1 > \frac 1{f(x)} > 0$ and if $f(x)$ grows blithely huge toward $huge$, $double-huge$ and $huge^2$ then $\frac 1{f(x)}$ will tend to flatline toward $0$.
So bearing that in mind:
$\cos x$ is periodic from $0$ to $1$ back down $0$ and and to $-1$ to $0$ to $1$ etc. and $\frac 1{\cos x}$ behaves exactly as we'd expect, for $+\infty$ down to a minimum of $1$ (as $\cos x$ goes from $0$ to $1$ and peaks) and then back to $\infty$.  Then as $\cos x$ goes from tiny positive through $0$, to tiny negative.  $\frac 1{\cos x}$ goess to $+\infty$ and jumps to $-\infty$ and starts surfaces up to $-1$ and then (as $\cos x$ reaches a nadir of $-1$ and heads back to $0$) $\frac 1{\cos x}$ reaches max at $-1$ and starts plumeting back to $-\infty.
As for $\tan x$ and $\frac 1{\tan x}$.  $\tan x$ will go from $\infty$ where $\cos x = 1; \sin x= 0$ to $1$ where $\cos x = \sin x = \frac 1{\sqrt 2}$. In this stage $\sin x$ is increasing faster than $\cos x$ is decreasing.  So there will be a bulge to the right in the graph.  Then as $\cos x\to 0$ and $\sin x \to 1$ we have $\frac {\sin x}{\cos x} \to \infty$. and as $\cos x$ is not decreasing faster than $\sin x$ is increasing, the graph stretches to be long and skinny.  Then as $\cos x$ goes through $0$ to negative, we jump to negative infinity.
In the same interval $\cot x = \frac {\cos x}{\sin x}$ starts and $\frac 10 = \infty$ and decreases down to $1$ and then to $0$.  The exact same but in reverse.  Then as $\cos x$ goes threough $0$ to negative $\frac {\cos x}{\sin x}$ goes ther $0$ to negative.
Useful to realize $\cot x =\frac {\cos x}{\sin x } = \frac {\sin(\frac \pi 2 - x)}{\cos \frac \pi 2 - x)} = \tan (\frac \pi 2 - x)$ so that explains more rigororously why their graphs are symmetriic.
A: Pictures often help.
Here's a graphical depiction of $\cos \theta$ and $\sec \theta = \frac{1}{\cos \theta}$ for $\theta = \frac{\pi}{3} = 60$ degrees.

The situation for $\cos \theta$ is represented by the right triangle $\triangle ABC$.  The radius is $CA = 1$, and the cosine of $\theta$ is represented by the length of $CB = \frac12$.
Now, extend $CA$ rightward to $A'$, which lies on the vertical line tangent to the unit circle.  I submit that the secant of $\theta$ is represented by the length $CA'$, on this basis:


*

*$\triangle ABC$ and $\triangle A'B'C$ are similar.

*Therefore $CA'$ is to $CB'$ as $CA$ is to $CB$.

*Since $CB' = CA = 1$, we have that $CA'$ is to $1$ as $1$ is to $CB$.

*Symbolically, $CA' = \frac{CA'}{1} = \frac{1}{CB}$.

*And since $CB$ is $\cos \theta$, $CA' = \frac{1}{\cos \theta} = \sec \theta$.


Finally, imagine increasing $\theta$ toward $\frac{\pi}{2} = 90$ degrees.  It is evident that $A'$ must move vertically upward—without bound, in fact, and that is indeed what the plot of $\sec \theta$ does.
If you increase $\theta$ beyond $\frac{\pi}{2}$, but continue to extend $AC$ rightward toward the same vertical tangent, you find that $\sec \theta$ is negative and initially large, but decreasing toward $0$ as you approach $\theta = \pi = 180$ degrees.
