---
tag: YDQ6P00V89
kind: proof
name: Proof of the addition formulas for hyperbolic functions
msc: [33B10]
proves: 0GJZCE87WP
uses: [0GWD21EWRW, 0EQFWKJ37J, 0RMVS7RXZX, 0E8HNVBCSP]
sources:
  - url: https://en.wikipedia.org/wiki/Hyperbolic_functions
    locator: Sums and differences of arguments
---

## Proof

We expand the right sides with the definitions and compare.

1. Let $x$ and $y$ be real numbers. Write $e^t = \exp(t)$ for the [[exponential function|0RMVS7RXZX]].
2. Let $p = e^x e^y$, $q = e^x e^{-y}$, $r = e^{-x} e^y$ and $s = e^{-x} e^{-y}$.
3. By the [[exponential of a sum|0E8HNVBCSP]], $p = e^{x + y}$ and $s = e^{-(x + y)}$.
4. By the definitions of the [[hyperbolic sine|0GWD21EWRW]] and the [[hyperbolic cosine|0EQFWKJ37J]], $\sinh t = \frac{e^t - e^{-t}}{2}$ and $\cosh t = \frac{e^t + e^{-t}}{2}$ for every real $t$.
5. By step 4, $4 \sinh x \cosh y = (e^x - e^{-x})(e^y + e^{-y}) = p + q - r - s$.
6. By step 4, $4 \cosh x \sinh y = (e^x + e^{-x})(e^y - e^{-y}) = p - q + r - s$.
7. Adding steps 5 and 6 gives $4(\sinh x \cosh y + \cosh x \sinh y) = 2p - 2s$.
8. By steps 3 and 4, $2p - 2s = 2(e^{x+y} - e^{-(x+y)}) = 4 \sinh(x + y)$. With step 7 this proves $\sinh(x + y) = \sinh x \cosh y + \cosh x \sinh y$.
9. By step 4, $4 \cosh x \cosh y = (e^x + e^{-x})(e^y + e^{-y}) = p + q + r + s$.
10. By step 4, $4 \sinh x \sinh y = (e^x - e^{-x})(e^y - e^{-y}) = p - q - r + s$.
11. Adding steps 9 and 10 gives $4(\cosh x \cosh y + \sinh x \sinh y) = 2p + 2s$.
12. By steps 3 and 4, $2p + 2s = 2(e^{x+y} + e^{-(x+y)}) = 4 \cosh(x + y)$. With step 11 this proves $\cosh(x + y) = \cosh x \cosh y + \sinh x \sinh y$.

## Idea

Each product of two hyperbolic functions expands into four products of exponentials. In the right combination, the mixed terms $e^{x - y}$ and $e^{y - x}$ cancel. The surviving terms $e^{x+y}$ and $e^{-(x+y)}$ assemble into the hyperbolic function of $x + y$.