Example 3.28 Using the above definitions, show that cosh^2x-sinh^2x=1 sinh(x+y)=coshxsinhy+sinhxcoshy 3. sinh(x+y)=sinhxcoshy+coshxsinhy

Let's correct and verify the given equations using the definitions of hyperbolic functions.

1. **Show that $\cosh^2(x) - \sinh^2(x) = 1$:**

By definition:
$$\cosh(x) = \frac{e^x + e^{-x}}{2}$$
$$\sinh(x) = \frac{e^x - e^{-x}}{2}$$

Squaring these expressions, we get:
$$\cosh^2(x) = \left( \frac{e^x + e^{-x}}{2} \right)^2 = \frac{e^{2x} + 2 + e^{-2x}}{4}$$
$$\sinh^2(x) = \left( \frac{e^x - e^{-x}}{2} \right)^2 = \frac{e^{2x} - 2 + e^{-2x}}{4}$$

Now, subtracting $\sinh^2(x)$ from $\cosh^2(x)$:
$$\cosh^2(x) - \sinh^2(x) = \frac{e^{2x} + 2 + e^{-2x}}{4} - \frac{e^{2x} - 2 + e^{-2x}}{4} = \frac{4}{4} = 1$$

Therefore, $\cosh^2(x) - \sinh^2(x) = 1$.

2. **Show that $\sinh(x+y) = \cosh(x)\sinh(y) + \sinh(x)\cosh(y)$:**

By definition:
$$\sinh(x+y) = \frac{e^{x+y} - e^{-(x+y)}}{2}$$

Using the exponential addition formula:
$$e^{x+y} = e^x e^y$$
$$e^{-(x+y)} = e^{-x} e^{-y}$$

Therefore:
$$\sinh(x+y) = \frac{e^x e^y - e^{-x} e^{-y}}{2} = \frac{e^x e^y - e^{-x} e^{-y}}{2} = \frac{e^x e^y - e^{-x} e^{-y}}{2}$$

Now, using the definitions of $\cosh(x)$ and $\sinh(y)$:
$$\cosh(x) = \frac{e^x + e^{-x}}{2}$$
$$\sinh(y) = \frac{e^y - e^{-y}}{2}$$

Similarly, for $\sinh(x)$ and $\cosh(y)$:
$$\sinh(x) = \frac{e^x - e^{-x}}{2}$$
$$\cosh(y) = \frac{e^y + e^{-y}}{2}$$

Now, combining these:
$$\cosh(x)\sinh(y) = \left( \frac{e^x + e^{-x}}{2} \right) \left( \frac{e^y - e^{-y}}{2} \right)$$
$$\sinh(x)\cosh(y) = \left( \frac{e^x - e^{-x}}{2} \right) \left( \frac{e^y + e^{-y}}{2} \right)$$

Adding these two results:
$$\cosh(x)\sinh(y) + \sinh(x)\cosh(y) = \frac{(e^x + e^{-x})(e^y - e^{-y}) + (e^x - e^{-x})(e^y + e^{-y})}{4}$$

Simplifying the numerator:
$$(e^x e^y - e^x e^{-y} + e^{-x} e^y - e^{-x} e^{-y}) + (e^x e^y + e^x e^{-y} - e^{-x} e^y - e^{-x} e^{-y})$$
$$= e^{x+y} - e^{-x-y} + e^{x-y} - e^{-x+y}$$

Combining like terms:
\[
= e^{x+y} - e^{-x-y}