Hyperbolic Sine And Cosines

The graphs of hyperbolic sine and cosine can be plotted using a graphing utility or by creating a table of values. The graph of $y = \sinh x$ can be visualized by graphing $y = \frac{1}{2}e^x$ and $y = -\frac{1}{2}e^{-x}$ separately and then adding the corresponding y-values (ordinates) as shown in Figure below

Similarly, the graph of $y = \cosh x$ can be obtained by graphing $y = \frac{1}{2}e^x$ and $y = \frac{1}{2}e^{-x}$ separately and then adding the corresponding ordinates

From the graphs:

• $\sinh x$ has a domain of $(-\infty, \infty)$ and a range of $(-\infty, \infty)$. It is an odd function (symmetric about the origin).
• $\cosh x$ has a domain of $(-\infty, \infty)$ and a range of $[1, \infty)$. It is an even function (symmetric about the y-axis).

From the definitions of $\sinh x$ and $\cosh x$, we can define four more hyperbolic functions:

1. Hyperbolic tangent: \tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}
2. Hyperbolic cotangent: \coth x = \frac{\cosh x}{\sinh x} = \frac{e^x + e^{-x}}{e^x - e^{-x}
3. Hyperbolic secant: \text{sech } x = \frac{1}{\cosh x} = \frac{2}{e^x + e^{-x}
4. Hyperbolic cosecant: \text{cosech } x = \frac{1}{\sinh x} = \frac{2}{e^x - e^{-x}

Graphs of these functions:

Graph of y = tanh x

Graph of y = coth x

Graph of y = sech x

Graph of y = cosech x

Many trigonometric identities have corresponding hyperbolic identities. Osborn's rule helps convert trigonometric identities to hyperbolic identities:

Osborn's rule: Replace trigonometric functions with their corresponding hyperbolic functions. Change the sign of any product of two sines.

Table of trigonometric and hyperbolic identities:

Trigonometric Identity	Corresponding Hyperbolic Identity
$\sin^2 x + \cos^2 x = 1$	$\cosh^2 x - \sinh^2 x = 1$
$1 + \tan^2 x = \sec^2 x$	$\cosh^2 x - \sinh^2 x=1$ which can be rearranged to $1 - \tanh^2 x = \text{sech}^2 x$
$1 + \cot^2 x = \csc^2 x$	$\coth^2 x - 1 = \text{cosech}^2 x$
$\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$	$\sinh(A \pm B) = \sinh A \cosh B \pm \cosh A \sinh B$
$\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$	$\cosh(A \pm B) = \cosh A \cosh B \pm \sinh A \sinh B$
$\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$	$\tanh(A \pm B) = \frac{\tanh A \pm \tanh B}{1 \pm \tanh A \tanh B}$
$\sin 2x = 2\sin x \cos x$	$\sinh 2x = 2\sinh x \cosh x$
$\cos 2x = \cos^2 x - \sin^2 x$	$\cosh 2x = \cosh^2 x + \sinh^2 x$
$\tan 2x = \frac{2\tan x}{1 - \tan^2 x}$	$\tanh 2x = \frac{2\tanh x}{1 + \tanh^2 x}$

Prove that $1 - \tanh^2 x = \text{sech}^2 x$.

Proof:

Recall the fundamental hyperbolic identity: $\cosh^2 x - \sinh^2 x = 1$.

Divide both sides by $\cosh^2 x$:

$\frac{\cosh^2 x}{\cosh^2 x} - \frac{\sinh^2 x}{\cosh^2 x} = \frac{1}{\cosh^2 x}$

$1 - \tanh^2 x = \text{sech}^2 x$

Prove that $\coth^2 x - 1 = \text{cosech}^2 x$.

Proof:

Recall the fundamental hyperbolic identity: $\cosh^2 x - \sinh^2 x = 1$.

Divide both sides by $\sinh^2 x$:

$\frac{\cosh^2 x}{\sinh^2 x} - \frac{\sinh^2 x}{\sinh^2 x} = \frac{1}{\sinh^2 x}$

$\coth^2 x - 1 = \text{cosech}^2 x$

Solve for $x$ in the equation $6\text{sech}^2 x + 7\tanh x - 8 = 0$.

Solution:

Using the identity $\text{sech}^2 x = 1 - \tanh^2 x$, we have:

$6(1 - \tanh^2 x) + 7\tanh x - 8 = 0$

$6 - 6\tanh^2 x + 7\tanh x - 8 = 0$

$-6\tanh^2 x + 7\tanh x - 2 = 0$

$6\tanh^2 x - 7\tanh x + 2 = 0$

Let $y = \tanh x$. Then $6y^2 - 7y + 2 = 0$.

$(2y - 1)(3y - 2) = 0$

$y = \frac{1}{2}$ or $y = \frac{2}{3}$

So, $\tanh x = \frac{1}{2}$ or $\tanh x = \frac{2}{3}$.

Using the inverse hyperbolic tangent function:

$x = \tanh^{-1}\left(\frac{1}{2}\right)$ or $x = \tanh^{-1}\left(\frac{2}{3}\right)$

Recall that $\tanh^{-1} u = \frac{1}{2}\ln\left(\frac{1+u}{1-u}\right)$. Therefore:

$x = \frac{1}{2}\ln\left(\frac{1 + \frac{1}{2}}{1 - \frac{1}{2}}\right) = \frac{1}{2}\ln 3 \quad \text{or} \quad x = \frac{1}{2}\ln\left(\frac{1 + \frac{2}{3}}{1 - \frac{2}{3}}\right) = \frac{1}{2}\ln 5$

Given the function $f(x) = 2\cosh^2 x + 3\sinh x = k$, where $k$ is a constant.

(a) Find the range of values of $k$ for which the equation has no real solutions. (b) If $k = 1$, find the exact solution of the equation.

Solution:

(a) Using $\cosh^2 x = 1 + \sinh^2 x$, we have:

$2(1 + \sinh^2 x) + 3\sinh x = k$

$2 + 2\sinh^2 x + 3\sinh x - k = 0$

$2\sinh^2 x + 3\sinh x + (2 - k) = 0$

Let $y = \sinh x$. Then $2y^2 + 3y + (2 - k) = 0$.

For no real solutions, the discriminant must be negative: $b^2 - 4ac < 0$.

$3^2 - 4(2)(2 - k) < 0$

$9 - 16 + 8k < 0$

$8k - 7 < 0$

$8k < 7$

$k < \frac{7}{8}$

(b) If $k = 1$, the equation becomes $2\sinh^2 x + 3\sinh x + 1 = 0$.

$(2\sinh x + 1)(\sinh x + 1) = 0$

$\sinh x = -\frac{1}{2}$ or $\sinh x = -1$

$x = \sinh^{-1}\left(-\frac{1}{2}\right)$ or $x = \sinh^{-1}(-1)$

Using $\sinh^{-1} u = \ln(u + \sqrt{u^2 + 1})$:

$x = \ln\left(-\frac{1}{2} + \sqrt{\frac{1}{4} + 1}\right) = \ln\left(\frac{-1 + \sqrt{5}}{2}\right) \quad \text{or} \quad x = \ln(-1 + \sqrt{1 + 1}) = \ln(-1 + \sqrt{2})$

Let $f(x) = 3 - \cosh x$, $x \in \mathbb{R}$.

(a) Sketch the graph of $f(x)$. Include the coordinates of any intercepts. (b) If $g(x) = \sinh x$, find the exact coordinates of the intersection points of $f(x)$ and $g(x)$.

Solution:

(a) x-intercepts: $3 - \cosh x = 0 \Rightarrow \cosh x = 3 \Rightarrow x = \pm \cosh^{-1} 3 = \pm \ln(3 + \sqrt{8}) = \pm \ln(3+2\sqrt{2})$. y-intercept: $f(0) = 3 - \cosh 0 = 3 - 1 = 2$. The graph is a reflection of cosh x about the x-axis, shifted up by 3 units.

(b) Intersection points: $3 - \cosh x = \sinh x \Rightarrow 3 = \sinh x + \cosh x \Rightarrow 3 = e^x \Rightarrow x = \ln 3$. Then $y = \sinh(\ln 3) = \frac{e^{\ln 3} - e^{-\ln 3}}{2} = \frac{3 - \frac{1}{3}}{2} = \frac{4}{3}$. Intersection point: $(\ln 3, \frac{4}{3})$.

A function has an inverse function if and only if it is one-to-one. Let's examine the inverse hyperbolic functions:

Inverse hyperbolic sine ($\sinh^{-1} x$ or $\text{arcsinh } x$)

$y = \sinh x$ is a one-to-one function, so its inverse exists. The graph of $y = \sinh x$ and its inverse $y = \sinh^{-1} x$ are reflections of each other across the line $y = x$

Domain of $\sinh x$: $(-\infty, \infty)$

Range of $\sinh x$: $(-\infty, \infty)$

Domain of $\sinh^{-1} x$: $(-\infty, \infty)$

Range of $\sinh^{-1} x$: $(-\infty, \infty)$

Inverse hyperbolic cosine ($\cosh^{-1} x$ or $\text{arccosh } x$)

$y = \cosh x$ is not one-to-one over its entire domain $(-\infty, \infty)$. However, if we restrict the domain of $\cosh x$ to $[0, \infty)$, it becomes one-to-one and has an inverse.

The graph of $y = \cosh x$ (for $x \ge 0$) and its inverse $y = \cosh^{-1} x$ are reflections of each other across the line $y = x$

Domain of $\cosh x$ (restricted): $[0, \infty)$

Range of $\cosh x$: $[1, \infty)$

Domain of $\cosh^{-1} x$: $[1, \infty)$

Range of $\cosh^{-1} x$: $[0, \infty)$

Inverse hyperbolic tangent ($\tanh^{-1} x$ or $\text{arctanh } x$)

$y = \tanh x$ is a one-to-one function, so its inverse exists. The graph of $y = \tanh x$ and its inverse $y = \tanh^{-1} x$ are reflections of each other across the line $y = x$

Domain of $\tanh x$: $(-\infty, \infty)$

Range of $\tanh x$: $(-1, 1)$

Domain of $\tanh^{-1} x$: $(-1, 1)$

Range of $\tanh^{-1} x$: $(-\infty, \infty)$

Inverse hyperbolic secant, cotangent, and cosecant

The inverse hyperbolic secant, cotangent, and cosecant are denoted as $\text{sech}^{-1} x$, $\coth^{-1} x$, and $\text{cosech}^{-1} x$, respectively. Their domains and ranges are derived from the domains and ranges of their corresponding hyperbolic functions, considering appropriate restrictions to ensure one-to-one mapping where necessary. They can be expressed in logarithmic form as follows:

$\text{sech}^{-1}(x) = \ln\left(\frac{1}{x} + \sqrt{\frac{1}{x^2} - 1}\right)$

$\coth^{-1}(x) = \frac{1}{2}\ln\left(\frac{x+1}{x-1}\right)$

$\text{cosech}^{-1}(x) = \ln\left(\frac{1}{x} + \sqrt{\frac{1}{x^2} + 1}\right)$

Inverse hyperbolic functions can be expressed in logarithmic form. This is achieved by expressing the hyperbolic function in terms of exponentials and then solving for the independent variable.

Example 1

Convert the following to logarithmic form:

(a) $y = \sinh^{-1} x$ (b) $y = \cosh^{-1} x$

Solution:

(a) $y = \sinh^{-1} x$

$x = \sinh y = \frac{e^y - e^{-y}}{2}$

$2x = e^y - \frac{1}{e^y}$

Multiply by $e^y$:

$2xe^y = e^{2y} - 1$

$e^{2y} - 2xe^y - 1 = 0$

This is a quadratic equation in $e^y$. Using the quadratic formula:

$e^y = \frac{-(-2x) \pm \sqrt{(-2x)^2 - 4(1)(-1)}}{2(1)} = \frac{2x \pm \sqrt{4x^2 + 4}}{2} = \frac{2x \pm 2\sqrt{x^2 + 1}}{2} = x \pm \sqrt{x^2 + 1}$

Since $e^y$ is always positive, and since $\sqrt{x^2 + 1} > |x|$, the only valid solution is the positive root:

$e^y = x + \sqrt{x^2 + 1}$

Taking the natural logarithm of both sides:

$y = \ln(x + \sqrt{x^2 + 1})$

Therefore, $\sinh^{-1} x = \ln(x + \sqrt{x^2 + 1})$

(b) $y = \cosh^{-1} x$

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

$2x = e^y + \frac{1}{e^y}$

Multiply by $e^y$:

$2xe^y = e^{2y} + 1$

$e^{2y} - 2xe^y + 1 = 0$

Using the quadratic formula:

$e^y = \frac{-(-2x) \pm \sqrt{(-2x)^2 - 4(1)(1)}}{2(1)} = \frac{2x \pm \sqrt{4x^2 - 4}}{2} = x \pm \sqrt{x^2 - 1}$

Since the range of $\cosh^{-1} x$ is $y \ge 0$, and the domain is $x \ge 1$, both solutions are mathematically valid, but by convention we use the positive root:

$y = \ln(x + \sqrt{x^2 - 1})$

Therefore, $\cosh^{-1} x = \ln(x + \sqrt{x^2 - 1})$, for $x \ge 1$.

Logarithmic forms of other inverse hyperbolic functions

Using similar methods, we can derive the logarithmic forms of the other inverse hyperbolic functions:

• $\tanh^{-1} x = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$, for $-1 < x < 1$
• $\coth^{-1} x = \frac{1}{2}\ln\left(\frac{x+1}{x-1}\right)$, for $|x| > 1$
• $\text{sech}^{-1} x = \ln\left(\frac{1 + \sqrt{1-x^2}}{x}\right)$, for $0 < x \le 1$
• $\text{cosech}^{-1} x = \ln\left(\frac{1}{x} + \frac{\sqrt{1+x^2}}{|x|}\right)$, for $x \ne 0$

Example 2

Approximate the following to four decimal places:

(a) $\sinh^{-1} 1$ (b) $\cosh^{-1} 3$ (c) $\coth^{-1} 4$

Solution:

(a) $\sinh^{-1} 1 = \ln(1 + \sqrt{1^2 + 1}) = \ln(1 + \sqrt{2}) \approx 0.8814$

(b) $\cosh^{-1} 3 = \ln(3 + \sqrt{3^2 - 1}) = \ln(3 + \sqrt{8}) = \ln(3 + 2\sqrt{2}) \approx 1.7627$

(c) $\coth^{-1} 4 = \frac{1}{2}\ln\left(\frac{4+1}{4-1}\right) = \frac{1}{2}\ln\left(\frac{5}{3}\right) \approx 0.2554$

Example 3

If $\coth^{-1} \frac{m}{2} - \coth^{-1} \frac{n}{2} = \frac{1}{2} \ln 8$, prove that $\frac{m+n}{m-n} = \frac{9}{7}$.

Solution:

$\coth^{-1} \frac{m}{2} - \coth^{-1} \frac{n}{2} = \frac{1}{2}\ln\left(\frac{\frac{m}{2}+1}{\frac{m}{2}-1}\right) - \frac{1}{2}\ln\left(\frac{\frac{n}{2}+1}{\frac{n}{2}-1}\right) = \frac{1}{2}\ln\left(\frac{m+2}{m-2}\right) - \frac{1}{2}\ln\left(\frac{n+2}{n-2}\right)$

$= \frac{1}{2} \ln\left(\frac{(m+2)(n-2)}{(m-2)(n+2)}\right)$

Given that this equals $\frac{1}{2} \ln 8$, we have:

$\frac{(m+2)(n-2)}{(m-2)(n+2)} = 8$

$mn - 2m + 2n - 4 = 8(mn + 2m - 2n - 4)$

$mn - 2m + 2n - 4 = 8mn + 16m - 16n - 32$

$7mn + 18m - 18n - 28 = 0$

This does not directly simplify to the requested $\frac{m+n}{m-n} = \frac{9}{7}$. There may be an error in the original problem statement or the intended solution. Further information is required to reach that conclusion.