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# Cosh|Definition & Meaning

**Definition**

The **hyperbolic cosine function**, or **cosh(x)**, is one of the various hyperbolic functions. Its evaluation involves **Euler’s number** **e**. For an input **x**, the hyperbolic cosine’s output is the **sum** of e to the power x and e to the power minus x, **divided** by 2. Unlike the trigonometric function cosine, which is based on measurements of **circles**, the hyperbolic cosine is based on measurements of **hyperbolas**.

In applied mathematics, **hyperbolic** **functions** are equivalents of regular **trigonometric functions**Â but expressed by means of the **hyperbola** instead of a **circle**.Â

Figure 1 – Graph of cosh x and sech x

Here we are going to look into the cosh (x), which is defined by the **formula** given below:

\[\cosh(x) = \dfrac{e^x + e^{(-x)}}{2} \]

In applied mathematics, the **exponential** **functions** are used in a variety of even and odd combinations that deserve to be given their own names. These **hyperbolic** **functions** are similar in many ways to the **trigonometric** **functions** because of their association with the **hyperbola** in the form of a **circle**.

Because of this, they are communally termed **hyperbolic** **functions** and separately termed **hyperbolic** **sine**, **hyperbolic** **cosine**, and so forth.

## What Are Trigonometric Functions?

Looking at a **right-angled** **triangle**, the side that is opposite to the right angle is known as the **hypotenuse,** and if we choose one other **angle** and name it Î¸ **theta**, then the other remaining sides are usually labeled as the **opposite** side, and the **adjacent** side (the side next to Î¸). The ratio of any two sides, like **adjacent**/ **hypotenuse,** will constitute a function with a special name, the **cosine** **of** **theta** or **cos** **(Î¸).**

Cos(Î¸) **=** $\dfrac{\textsf{adjacent}}{\textsf{hypotenuse}}$

Similarly, considering the other possible ratios, we can construct the other **trigonometric** **functions:**

Sin(Î¸) **=** $\dfrac{\textsf{opposite}}{\textsf{hypotenuse}}$

Tan(Î¸) **= **$\dfrac{\textsf{opposite}}{\textsf{adjacent}}$

The **cosine** can be defined using a **visual** **representation** by considering a circle of **diameter** 2, having center **O **at the origin inside the (x, y) **plane**. Consider **point** **A** on the **edge** of the **circle** having the **coordinates** (1, 0). Now the distance from that point to the **origin** is called as **OA **and has a length of **1 unit**.

Let there be another **point** **P** on the line of the circle’s arbitrary **coordinates** (a, b). The angle between the two lengths, **OA and OP,** is defined by **Î¸.** Here theta is the radian representation.

In simpler terms, the cosine of Î¸ is described as the projection of P on the x-axis.

## What Are Hyperbolic Functions?

**In applied mathematics, hyperbolic functions are equivalents to regular trigonometric functions but expressed by means of the hyperbola instead of a circle.** While the points (cos t, sin t) are derived from a circle having a **unit radius**, the functions (cosh t, sinh t) develop the rightmost half of the **unit** **hyperbola**.

Their derivatives are also similar to the derivatives of trigonometric functions, except the sign. We know the derivatives of **sin(t) **and** cos(t) **are, respectively,** cos(t) **and** â€“sin(t), **while the derivatives of **sinh(t)** and **cosh(t) **are** cosh(t) **and** +sinh(t)**.

**Hyperbolic functions** arise where one needs to compute **angles and distances** in **hyperbolic** **geometry**. They are also used in various solutions of linear **differential** **equations**, **cubic** **functions**, and **Laplace’s** **equation**. **Laplace’s** **equations** are significant in numerous fields of physics, which include **electromagnetic concepts, heat fusion, etc.**

All the **hyperbolic** **functions** acquire a real value termed a **hyperbolic** **angle**. The **hyperbolic** **angle** is double the **size** of the **area** of the **hyperbolic** **segment** it is constructed in.

## Defining Hyperbolic Cosine

We can draft the **graph for hyperbolic cosine** by utilizing the given information of the exponential $e^x$ and $e^{âˆ’x}$. Initially, let us compute the value of **cosh 0** by taking **x = 0. **Thus, the exponents become $e^x$** = 1 and **$e^{âˆ’x}$** = 1.** Thus:

\[\cosh(x) = \dfrac{e^0 + e^{(-0)}}{2} \]

= (1 + 1) / 2 = 1

Rewriting the **cosh** **x** to determine the change when the value of **x** gets large:

\[\cosh(x) = \dfrac{e^{x/2} + e^{x/2}}{2} \]

The **graph** below shows the **behavior** of the **exponential** when **x** gets large.

Looking at the graph, the positive **exponential** increases quickly when **x** gets larger, but the **negative** **exponential** **decreases** rapidly with the change in x. So as the x gets larger, the **hyperbolic** **cosine** gets closer and closer to the **positive** **exponential,** i.e., $e^{x/2}$:

$\cosh x = e^{x/2}$Â Â for larger values of x

The **graph** of cosh x will **eventually remain above** the graph of the $e^{x/2}$ due to the fact that the second part of the function, despite being very small, will always be greater than zero. **As the value of x gets bigger and bigger, the difference in both graphs gets minor and minor.**

Now, Letâ€™s say that the value of x becomes **more and more negative**, the **positive exponential will fall drastically**, but the **negative exponential increases rapidly** with the negative change in x. So as the x gets smaller and smaller, the **hyperbolic cosine gets closer and closer to the negative exponential, **i.e., $e^{-x/2}$:

$\cosh x = e^{-x/2}$Â Â for large negative values of x

Even if the x is negative, the graph of cosh x will **ultimately remain above** the graph of the $e^{-x/2}$ because the first part of the sum, regardless of being very small, will always be **greater than zero**. As the value of **x **gets smaller and smaller, the difference in both graphs gets **negligible**.

Thus, we can construct the **graph** for **cosh** **x**. One thing to observe is that the **graph** is **symmetrical** about the **y-axis** because **cosh x = cosh (-x),** as described above.

## Important Identity of cosh

**Cosh,** along with **sinh,** have various identities that look analogous to **identities** for the regular **trigonometric** **functions** of **cos** and **sin**, with a slight change in the **signs**. The identity looks like this:

\[\cosh^{2} x-\sinh^{2} x = 1\]

We can recall the **trigonometric** **identity** similar to the one above $\cos^2 x + \sin^2 x = 1$, with the **plus** **sign** changing to **minus**.

The identity given above can **effortlessly** be **drawn** from the **fundamental** **definitions** as follows:

\[\cosh^{2} x-\sinh^{2} x = \dfrac{e^x + e^{(-x)}}{2}-\dfrac{e^x – e^{(-x)}}{2}\]

\[= \dfrac{e^{2x} + e^{-2x}}{4}-\dfrac{e^{2x}-e^{(-2x)}}{4}\]

\[= \frac{1}{2}-\frac{-1}{2}\]

\[= 1\]

Another **important** **identity** to look for is:

**Sinh(x + y) = sinh x cosh y + cosh x sinh y**

**Osbornâ€™s Rule**

**Osbornâ€™s rule** implies that the **trigonometric** **identities** can be **converted** to their corresponding **hyperbolic** **identities** by replacing all **occurrences** of **sins** and **cosine** with their counter **hyperbolic** **sine** and **cosine**. Also, if them occurs a product of two sines, the corresponding hyperbolic product will have a **negative** **sign**.

For instance, $\cos(2A) = 2\cos^2(A)-1$ will be changed into $\cosh2A = 2\cosh^2(A)-1$ whereas,

\[\cos(2A) = 1-\sin^2(A) \Rightarrow \cosh(2A) = 1+\sinh^2(A)\]

In order to apply **Osbornâ€™s rule** on other **trigonometric** **identities**, they must first be narrowed down to **sines** and **cosines**.

**Solved Example**

For the equation **5 cosh x + 3 sinh x = 4,** solve for x.

**Solution**

Using the basic definitions of cosh and sinh, the given equation can be written as:

\[5 \dfrac{e^x + e^{(-x)}}{2} + 3 \dfrac{e^x + e^{(-x)}}{2} = 4\]

Simplifying it gives us the following:

\[4e^x + e^{-x} = 4\]

Multiplying by $e^x$ gives:

\[4e^{2x} + 1 = 4e^x\]

Rewriting this forms a **quadratic** **equation**:

\[4e^{2x}-4e^x + 1 = 0\]

Solving the **quadratic** **equation** gives:

\[e^x = \dfrac{1}{8} (4 \pm \sqrt{0})\]

\[e^x = \dfrac{1}{2}\]

Hence:

\[x = \ln\left(\frac{1}{2}\right)\]

\[= \ln 2^{-1}\]

\[= -\ln2\]

*All images/mathematical drawings were created with GeoGebra.*