$\cos x = \cosh x?$

Geometry Level 1

Billy Tangent naively thought that the hyperbolic cosine function and the standard cosine function were the same. To make sure, he tried one real value and, sure enough, he got the same result.

What value did he try?

If you think there are multiple values for which this would work, enter 99999 as your answer.
If you think there are no values for which this would work, enter 88888 as your answer.

The answer is 0.

1 solution

Geoff Pilling
Jan 27, 2017

$\cosh x = \dfrac{e^x + e^{-x}}{2}$

This function is strictly greater that one except at the point $(0,1)$ for which it equals $1$ .

And, $-1 \leq \cos x \leq 1$

Therefore, the only point for which the two will equal is at the point $(0,1)$ .

So, $x = \boxed{0}$

Interesting follow-ups: the same question instead with $\sin(x) = \sinh(x)$ and $\sec(x) = \cosh(x)$ . For the former the answer would still be just $x = 0$ but the behavior of the two curves near the origin would have to be analyzed using calculus to verify. For the latter, if we restrict ourselves to $(-\frac{\pi}{2}, \frac{\pi}{2})$ then the answer would still be $x = 0$ , (as $\sec(x)$ is "more concave up" on this interval), but over all reals there would be an infinite number of points of intersection.

Brian Charlesworth - 4 years, 4 months ago

Another interesting one is $\csc x = \text{csch} x$ as its not quite so obvious just looking at the graphs where the intersections are...

Geoff Pilling - 4 years, 4 months ago

Yeah, that one is tricky. I think that there are 4 points of intersection, but they're within a whisker of each other as $x \to 0$ from either side.

Brian Charlesworth - 4 years, 4 months ago

@Brian Charlesworth – Whoa... Whisker is right!

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

Geoff Pilling - 4 years, 4 months ago

I think x=0 is the only solution even for complex values.

Ishwar Karthik - 3 years, 8 months ago

cos ⁡ x = cosh ⁡ x ? \cos x = \cosh x? cos x = cosh x ?

The answer is 0.

1 solution

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$\cos x = \cosh x?$