# Thread: Intersecting reciprocal trig functions

1. ## Intersecting reciprocal trig functions

Hi,

I hope someone can help. I'm trying to understand intuitively why y=cscx intersects y=secx at (pi/4, 1.414) and (pi/4, -1.414)? Can someone please explain why?

I noticed that the parent functions of both reciprocals intersect at (pi/4, 0.707) and (pi/4, -0.707). I also noticed that 1.414 = 0.707 x 2. So the x-values of their intersection points are the same, and the y-values of their intersection points are double. I don't fully know what to make of this analysis of mine, however.

I would appreciate some enlightenment on this.

Sincerely,
Olivia

2. ## Re: Intersecting reciprocal trig functions

You want to find values of $x$ such that:
$\sec x = \csc x$

$\sec x = \dfrac{1}{\cos x} = \dfrac{1}{\sin x} = \csc x$

We can focus on the middle equality:

$\dfrac{1}{\cos x} = \dfrac{1}{\sin x}$

Cross multiply
$\sin x = \cos x$

So, the intersect at all values of $x$ where $\sin x$ intersects $\cos x$.

You already said that those intersect at $x$-value $\dfrac{\pi}{4}$. Let's plug that into our $\sec x$ and $\csc x$ functions:

$\sec \left( \dfrac{\pi}{4} \right) = \dfrac{1}{ \cos \left( \dfrac{\pi}{4} \right) } = \dfrac{1}{ \left( \dfrac{1}{ \sqrt{2} } \right)} = \sqrt{2}$
$\csc \left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sin \left(\dfrac{\pi}{4}\right)} = \dfrac{1}{\left(\dfrac{1}{\sqrt{2}}\right)} = \sqrt{2}$

3. ## Re: Intersecting reciprocal trig functions

Originally Posted by otownsend
I'm trying to understand intuitively why y=cscx intersects y=secx at (pi/4, 1.414) and (pi/4, -1.414)? Can someone please explain why?
Because \begin{align*}\sin \left( {\frac{\pi }{4}} \right) = \cos \left( {\frac{\pi }{4}} \right) &= \frac{{\sqrt 2 }}{2}\\\csc\left( {\frac{\pi }{4}} \right)=\sec\left( {\frac{\pi }{4}} \right)&=\frac{2}{\sqrt{2}}\end{align*}

NOW you must know that (\forall n\in\mathbb{Z})\left[ \begin{align*}\sin \left( {\frac{\pi }{4}+n\pi} \right) = \cos \left( {\frac{\pi }{4}+n\pi} \right) &= \frac{{\sqrt 2 }}{2}\\\csc\left( {\frac{\pi }{4}+n\pi} \right)=\sec\left( {\frac{\pi }{4}+n\pi} \right)&=\frac{2}{\sqrt{2}}\end{align*}\right]

4. ## Re: Intersecting reciprocal trig functions

Originally Posted by Plato
Because \begin{align*}\sin \left( {\frac{\pi }{4}} \right) = \cos \left( {\frac{\pi }{4}} \right) &= \frac{{\sqrt 2 }}{2}\\\csc\left( {\frac{\pi }{4}} \right)=\sec\left( {\frac{\pi }{4}} \right)&=\frac{2}{\sqrt{2}}\end{align*}

NOW you must know that (\forall n\in\mathbb{Z})\left[ \begin{align*}\sin \left( {\frac{\pi }{4}+n\pi} \right) = \cos \left( {\frac{\pi }{4}+n\pi} \right) &=(-1)^n \frac{{\sqrt 2 }}{2}\\\csc\left( {\frac{\pi }{4}+n\pi} \right)=\sec\left( {\frac{\pi }{4}+n\pi} \right)&= (-1)^n \frac{2}{\sqrt{2}}\end{align*}\right]
Small typo. You missed the negative sign when $n$ is odd. I updated it in the quote.

5. ## Re: Intersecting reciprocal trig functions

I understand it. I think the best way to summarize this is as follows: The reason why sin and cos intersect at these points in the first place is because, looking back at the special triangles, the only triangle that has the same opposite and adjacent sides is the 45-45-90 triangle. This results in the ratio between sides for both sine and cosine to be the same. And since they are the same, this means that there reciprocals will also be the same. When looking at a graph, we always have to bring it back to the special triangles in order to understand the behaviour of why the graph is the way it is.