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To prove the identity for hyperbolic functions:

\[
\cosh(x + y) = \cosh(x) \cosh(y) + \sinh(x) \sinh(y),
\]

we can use the definitions of the hyperbolic sine and cosine functions:

\[
\cosh(z) = \frac{e^z + e^{-z}}{2}, \quad \sinh(z) = \frac{e^z - e^{-z}}{2}.
\]

### Proof:

Start with the left-hand side:

\[
\cosh(x + y) = \frac{e^{x+y} + e^{-(x+y)}}{2}.
\]

Now expand this expression:

\[
\cosh(x + y) = \frac{e^x e^y + e^{-x} e^{-y}}{2}.
\]

Now, let's express the right-hand side of the identity we want to prove:

\[
\cosh(x) \cosh(y) + \sinh(x) \sinh(y).
\]

Using the definitions of \(\cosh\) and \(\sinh\), we substitute:

\[
\cosh(x) \cosh(y) = \frac{e^x + e^{-x}}{2} \cdot \frac{e^y + e^{-y}}{2} = \frac{(e^x e^y + e^x e^{-y} + e^{-x} e^y + e^{-x} e^{-y})}{4},
\]

and

\[
\sinh(x) \sinh(y) = \frac{e^x - e^{-x}}{2} \cdot \frac{e^y - e^{-y}}{2} = \frac{(e^x e^y - e^x e^{-y} - e^{-x} e^y + e^{-x} e^{-y})}{4}.
\]

Now, add the two expressions:

\[
\cosh(x) \cosh(y) + \sinh(x) \sinh(y) = \frac{e^x e^y + e^x e^{-y} + e^{-x} e^y + e^{-x} e^{-y}}{4} + \frac{e^x e^y - e^x e^{-y} - e^{-x} e^y + e^{-x} e^{-y}}{4}.
\]

Combining terms:

\[
= \frac{(e^x e^y + e^{-x} e^{-y}) + (e^x e^y + e^{-x} e^{-y})}{4}
= \frac{2(e^x e^y + e^{-x} e^{-y})}{4}
= \frac{e^x e^y + e^{-x} e^{-y}}{2}.
\]

This is exactly the same expression as the left-hand side \(\cosh(x + y)\).

Thus, we have shown that:

\[
\cosh(x + y) = \cosh(x) \cosh(y) + \sinh(x) \sinh(y).
\]

Would you like any further explanation or clarification? 

Here are some related questions for deeper exploration:

1. What are the definitions of hyperbolic sine and cosine functions?
2. How does the hyperbolic cosine function relate to exponential functions?
3. Can you prove a similar identity for \(\sinh(x + y)\)?
4. How do hyperbolic identities compare to trigonometric identities?
5. What are the applications of hyperbolic functions in physics or engineering?

**Tip:** The hyperbolic identities are closely analogous to the trigonometric addition formulas, with \(\cosh(x)\) and \(\sinh(x)\) playing similar roles to \(\cos(x)\) and \(\sin(x)\), respectively.

Prove that cosh(x + y) = cosh(x)cosh(y) + sinh(x)sinh(y).

In this problem, we prove the identity cosh(x + y) = cosh(x)cosh(y) + sinh(x)sinh(y) using hyperbolic function definitions. Detailed step-by-step solution provided, ideal for Grades 11-12 and university students.

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