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.

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(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.

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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.