Four intersections?

Geometry Level 4

Charley Brians was playing around with the following 6 trigonometric functions:

$\sin$
$\cos$
$\tan$
$\cot$
$\sec$
$\csc$ .

He noticed that, for one of them, if he sets it equal to its hyperbolic counterpart— $\sinh, \cosh, \tanh, \coth, \text{sech},$ or $\text{csch},$ respectively—it intersects at exactly four points.

Assuming that we are only dealing with real numbers, which trig function did he pick?

Inspiration: Brian Charlesworth

\sin

\cos

\tan

\cot

\sec

\csc

None of the 6 above More than one of the 6 above

1 solution

Geoff Pilling
Jan 27, 2017

Let, $\csc x = \text{csch } x$ for real $x$ .

Then, these graphs intersect at exactly two points:

$(-0.000102, -9804)$
$(-0.000022, -45455)$
$(0.000022, 45455)$
$(0.000102, 9804)$

All of the other trigonometric functions intersect with either one or an infinite number of points when set equal to their hyperbolic trigonometric counterpoints:

Function(s)	Intercepts
$\cos$	one $(0,1)$
$\sin$	one $(0,0)$
$\tan$	$\infty$
$\cot$	$\infty$
$\sec$	one $(0,1)$

So, the answer is $\boxed{\csc}$

Wow, those intersection points are pretty extreme! The series expansions for these two functions are

$\csc(x) = \dfrac{1}{x} + \dfrac{x}{6} + \dfrac{7x^{3}}{360} + \dfrac{31x^{5}}{15120} + ....$ and

$csch(x) = \dfrac{1}{x} - \dfrac{x}{6} + \dfrac{7x^{3}}{360} - \dfrac{31x^{5}}{15120} + ....$ ,

so no wonder they're so intertwined as $x \to 0$ !

Thanks for the acknowledgement; "Charley Brians" makes me think of "Charlie Brown", who I often totally identify with. :) One thought:- to bring the "More than one of these" option into play, you could start the second sentence with "He notices that for at least one of them, ...". The actual question could then be "Which trig function(s) did he pick?" The "None of these options" is in play for those who might think that Charley has miscalculated, (just like he does when he goes to kick the football that Lucy van Pelt has teed up).

Brian Charlesworth - 4 years, 4 months ago

Haha... Yup!

Although I don't have a cool name that sort of morphs into a well known cartoon character, I too can relate to Charlie Brown... Halloween special - "I got a rock!"

So, you think maybe a better question might be: Which of the following equations has more than one solution? Or maybe more than two? (Since they might guess that csc=csch has two but not four, since the intersections must be an even number?)

Geoff Pilling - 4 years, 4 months ago

Or perhaps, below the list of 6 six trig functions, "Which of the above functions has more than 2 but still a finite number of points of intersection with its hyperbolic counterpart, (sinh, cosh, tanh, coth, sech or csch, respectively)?"

I guess the title would have to be changed as well. Maybe "Would The Great Pumpkin Know?" :)

Brian Charlesworth - 4 years, 4 months ago

Do we are concedering fundamental period of inverse function

Dhruv Joshi - 4 years, 2 months ago

Your solution simply states the intersection points. Can you please give a description of your method of finding the points?

Atomsky Jahid - 4 years, 2 months ago

Yep! Is there a way to solve it numerically?

A Former Brilliant Member - 3 years, 5 months ago

Mine for tanh and coth. 2 is comming..which interval have you taken..?

Dhruv Joshi - 4 years, 2 months ago

I don't get it, if we have $csc(x)=csch(x)$ Then this is the same as $\frac{1}{sin(x)} = \frac{1}{sinh(x)}$ So $sin(x) = sinh(x)$ So why does $csc$ work but not $sin$ ?

Leonardo Finocchiaro - 2 years, 1 month ago

Four intersections?

1 solution

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