On The Shape Of Trig Graphs To A Precalc Student

Question

I want to know why 3 types of trig graphs have the shape that they do:

1. 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?

2. Cosecant graphs. Same as above.

3. 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.

Answer 1

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.

Answer 2

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.