Explaining sinh and cosh graphically

[Seung Tai Kang]

Homework Statement

tanh x is 12/13 . Then cosh x is 12/5 and sinh is 13/5. This can be found algebraically by finding x.

Homework Equations

tanh x= sinh x/cosh x

The Attempt at a Solution

The answer can be found algebraically by finding x. tanh x= sinh x /cosh x.

tanh(x) =
sinh(x) / cosh(x) =
((1/2) * (e^(x) - e^(-x))) / ((1/2) * (e^(x) + e^(-x))) =>
(e^(x) - e^(-x)) / (e^(x) + e^(-x))

12 / 13 = (e^(x) - e^(-x)) / (e^(x) + e^(-x))
12 * (e^(x) + e^(-x)) = 13 * (e^(x) - e^(-x))
12 * e^(x) + 12 * e^(-x) = 13 * e^(x) - 13 * e^(-x))
25 * e^(-x) = e^(x)
25 = e^(2x)
ln(25) = 2x
2 * ln(5) = 2x
x = ln(5)

sinh(x) =
(1/2) * (e^(x) - e^(-x)) =>
(1/2) * (5 - (1/5)) =>
(1/2) * (24/5) =>
12/5

cosh(x) =
(1/2) * (e^(x) + e^(-x)) =
(1/2) * (26/5) =>
13/5

But I believe there must be a graphical explanation using hyperbola too, just as there are algebraic and graphic explanations for trigonometry using circles and triangle.

Can someone explain it to me graphically using hyperbola?

 

[Orodruin]

It is just the same as the circle, but using the unit hyperbola instead of the unit circle. Also, $x$ is not a regular angle, but a hyperbolic angle. The unit hyperbola can be parametrised by a hyperbolic angle $\theta$ as
$$
x = \cosh(\theta), \quad y = \sinh(\theta),
$$
since this implies that $x^2 - y^2 = 1$. Draw a line through the origin and $x = 13, y = 12$. Where it crosses the hyperbola, the $x$-value is $\cosh(\theta)$ and the $y$-value is $\sinh(\theta)$. Of course, to do this you will need to draw the hyperbola properly first.

However, your problem is more easily solved using the hyperbolic identity $\cosh^2(\theta) - \sinh^2(\theta) = 1$ to relate $\cosh(\theta)$ and $\sinh(\theta)$ to $\tanh(\theta)$. You do not need to go via the definitions in terms of exponential functions.

 

[Seung Tai Kang]

It is just the same as the circle, but using the unit hyperbola instead of the unit circle. Also, $x$ is not a regular angle, but a hyperbolic angle. The unit hyperbola can be parametrised by a hyperbolic angle $\theta$ as
$$
x = \cosh(\theta), \quad y = \sinh(\theta),
$$
since this implies that $x^2 - y^2 = 1$. Draw a line through the origin and $x = 13, y = 12$. Where it crosses the hyperbola, the $x$-value is $\cosh(\theta)$ and the $y$-value is $\sinh(\theta)$. Of course, to do this you will need to draw the hyperbola properly first.

However, your problem is more easily solved using the hyperbolic identity $\cosh^2(\theta) - \sinh^2(\theta) = 1$ to relate $\cosh(\theta)$ and $\sinh(\theta)$ to $\tanh(\theta)$. You do not need to go via the definitions in terms of exponential functions.

x=12 and y=13 does not cross the origin but just a straight line parallel to x and y-axis at x=12 and y=13 respectively. I guess you may have meant it with respect to θ but could you elaborate on it?

 

[Orodruin]

x=12 and y=13 does not cross the origin but just a straight line parallel to x and y-axis at x=12 and y=13 respectively. I guess you may have meant it with respect to θ but could you elaborate on it?

You have misread the post. There is only one line. It passes through the origin and the point $(x,y) = (12,13)$. Two points are sufficient to uniquely identify a line.