Prove the following double angle identities for the hyperbolic function.
(a) sinh2x = 2sinhxcoshx
(b) cosh2x = cosh2x+sinh2x

Let's prove the given hyperbolic double angle identities step by step:

### (a) \( \sinh(2x) = 2\sinh(x)\cosh(x) \)

**Proof:**
The hyperbolic sine function is defined as:

\[
\sinh(x) = \frac{e^x - e^{-x}}{2}
\]

The hyperbolic cosine function is defined as:

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

We want to show that \( \sinh(2x) = 2\sinh(x)\cosh(x) \).

Start with the definition of \( \sinh(2x) \):

\[
\sinh(2x) = \frac{e^{2x} - e^{-2x}}{2}
\]

Now, express \( e^{2x} \) and \( e^{-2x} \) using properties of exponents:

\[
\sinh(2x) = \frac{\left( e^x \right)^2 - \left( e^{-x} \right)^2}{2}
\]

This expression is a difference of squares, so we can factor it as:

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

But we know from the definitions of hyperbolic functions that:

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

So:

\[
\sinh(2x) = 2\sinh(x)\cosh(x)
\]

This proves the identity.

---

### (b) \( \cosh(2x) = \cosh^2(x) + \sinh^2(x) \)

**Proof:**
The hyperbolic cosine function \( \cosh(x) \) is:

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

We want to prove that:

\[
\cosh(2x) = \cosh^2(x) + \sinh^2(x)
\]

Start by expressing \( \cosh(2x) \) in terms of exponentials:

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

Next, expand \( e^{2x} \) and \( e^{-2x} \) in terms of \( e^x \) and \( e^{-x} \):

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

Now recall the definitions of \( \cosh(x) \) and \( \sinh(x) \):

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

We want to check if:

\[
\cosh(2x) = \cosh^2(x) + \sinh^2(x)
\]

Substitute the definitions of \( \cosh(x) \) and \( \sinh(x) \) into the right-hand side:

\[
\cosh^2(x) + \sinh^2(x) = \left( \frac{e^x + e^{-x}}{2} \right)^2 + \left( \frac{e^x - e^{-x}}{2} \right)^2
\]

Now expand both squares:

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

Using the identities \( (a + b)^2 = a^2 + 2ab + b^2 \) and \( (a - b)^2 = a^2 - 2ab + b^2 \), we get:

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

Now combine the terms:

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

This is exactly \( \cosh(2x) \), which proves the identity.

---

### Summary:
- (a) \( \sinh(2x) = 2\sinh(x)\cosh(x) \)
- (b) \( \cosh(2x) = \cosh^2(x) + \sinh^2(x) \)

Would you like further details or have any questions on these proofs?

### Relative Questions:
1. How do the hyperbolic identities differ from trigonometric identities?
2. Can we derive other double angle identities from these hyperbolic ones?
3. How are hyperbolic functions used in real-world applications?
4. What are the differences between the definitions of \( \sinh(x) \) and \( \cosh(x) \)?
5. Can we extend these identities to complex numbers?

### Tip:
The hyperbolic functions \( \sinh(x) \) and \( \cosh(x) \) are closely related to exponential functions, which makes them useful in solving differential equations, particularly in engineering and physics.

Explore detailed proofs of the hyperbolic double angle identities sinh(2x) = 2sinh(x)cosh(x) and cosh(2x) = cosh^2(x) + sinh^2(x). This solution covers key concepts in hyperbolic functions and their relationship with exponential functions, suitable for undergraduate mathematics students.

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