📂Functions

Sum and Difference Formulas and Multiplication Formulas for Hyperbolic Functions

Formulas

• Sum and Difference Formulas:

$$\begin{align} \sinh x +\sinh y =&\ 2\sinh \left(\frac{x+y}{2}\right) \cosh \left(\frac{x-y}{2}\right) \\[1em] \sinh x -\sinh y =&\ 2\sinh \left(\frac{x-y}{2}\right) \cosh \left( \frac{x+y}{2} \right) \\[1em] \cosh x + \cosh y =&\ 2 \cosh \left(\frac{x+y}{2}\right) \cosh \left(\frac{x-y}{2}\right) \\[1em] \cosh x -\cosh y =&\ 2 \sinh \left( \frac{x+y}{2} \right) \sinh \left(\frac{x-y}{2}\right) \end{align}$$

• Product Formulas:

$$\begin{align} \sinh x \sinh y =&\ \frac{\cosh (x+y)-\cosh (x-y)}{2} \\ \sinh x \cosh y =&\ \frac{\sinh (x+y)+\sinh (x-y)}{2} \\ \cosh x \sinh y =&\ \frac{\sinh (x+y)-\sinh (x-y)}{2} \\ \cosh x \cosh y =&\ \frac{\cosh (x+y)+\cosh (x-y)}{2} \end{align}$$

Description

The proof process is the same as the one used to derive the sum and difference formulas of trigonometric functions, so it will not be introduced in detail.

Proofs

Proof of $(1)-(4)$

According to the addition theorem,

$$\begin{align*} \sinh (x + y) =&\ \sinh x \cosh y + \sinh y \cosh x \\ \sinh (x - y) =&\ \sinh x \cosh y - \sinh y \cosh x \end{align*}$$

If we substitute $x=\frac{z+w}{2}$ and $y=\frac{z-w}{2}$, the above equation becomes

$$\begin{align*} \sinh z =&\ \sinh \frac{z+w}{2} \cosh \frac{z-w}{2} + \sinh \frac{z-w}{2} \cosh \frac{z+w}{2} \\ \sinh w =&\ \sinh \frac{z+w}{2} \cosh \frac{z-w}{2} - \sinh \frac{z-w}{2} \cosh \frac{z+w}{2} \end{align*}$$

If we add and subtract the above equation, we get the following, respectively.

$$\begin{align*} \sinh z+\sinh w =&\ 2\sinh \frac{z+w}{2} \cosh \frac{z-w}{2} \\ \sinh z -\sinh w =&\ 2\sinh \frac{z-w}{2} \cosh \frac{z+w}{2} \end{align*}$$

The rest can be obtained in the same way.

■

Proof of $(5)-(8)$

According to the addition theorem,

$$\begin{align*} \cosh (x + y) =&\ \cosh x \cosh y + \sinh x \sinh y \\ \cosh (x - y) =&\ \cosh x \cosh y - \sinh x \sinh y \end{align*}$$

Subtracting the below equation from the above,

$$\begin{align*} &&\cosh (x+y)-\cosh (x-y)=&\ 2\sinh x \sinh y \\ \implies && \sinh x \sinh y =&\ \frac{\sinh (x+y)+\sinh (x-y)}{2} \end{align*}$$

The rest can be obtained in the same way.

■