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# AI Hyperbolic Functions Solver

## About MathCrave AI Hyperbolic Solver

The AI hyperbolic solver is capable of solving hyperbolic functions, such as sinh x, cosh x, tanh x, cosech x, sech x, and coth x. It also provides a clear and detailed step-by-step worksheet including hyperbolic identities.

### Math Problems AI Hyperbolic Functions Solver Solves

• Sinh x, cosh x, tanh x, cosech x, sech x and coth x

• Evaluate hyperbolic functions

• State and proof Osborne’s rule

• Simple hyperbolic identities proofs

• Equations involving hyperbolic functions

• Derivatives the series expansions for cosh x

• Derivatives of the series expansions for sinh x

### Lesson Notes: Hyperbolic Functions

#### 1. Definitions of Hyperbolic Functions

Hyperbolic Sine (sinh x):
$\sinh x = \frac{e^x – e^{-x}}{2}$

Hyperbolic Cosine (cosh x):
$\cosh x = \frac{e^x + e^{-x}}{2}$

Hyperbolic Tangent (tanh x):
$\tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x – e^{-x}}{e^x + e^{-x}}$

Hyperbolic Cosecant (cosech x):
$\cosech x = \frac{1}{\sinh x} = \frac{2}{e^x – e^{-x}}$

Hyperbolic Secant (sech x):
$\sech x = \frac{1}{\cosh x} = \frac{2}{e^x + e^{-x}}$

Hyperbolic Cotangent (coth x):
$\coth x = \frac{\cosh x}{\sinh x} = \frac{e^x + e^{-x}}{e^x – e^{-x}}$

#### 2. Evaluate Hyperbolic Functions

Example: Evaluate $$\sinh 1$$ and $$\cosh 1$$

– $$\sinh 1 = \frac{e^1 – e^{-1}}{2} \approx \frac{2.71828 – 0.36788}{2} \approx 1.17520$$
– $$\cosh 1 = \frac{e^1 + e^{-1}}{2} \approx \frac{2.71828 + 0.36788}{2} \approx 1.54308$$

#### 3. Osborne’s Rule

Statement: Osborne’s rule states that to convert a trigonometric identity to a hyperbolic identity, change the trigonometric functions to hyperbolic functions and change the sign of any product of two sines.

Proof: Consider the trigonometric identity $$\cos^2 x – \sin^2 x = 1$$.

– Replace $$\cos x$$ with $$\cosh x$$ and $$\sin x$$ with $$\sinh x$$:
$\cosh^2 x – \sinh^2 x = 1$

– This is a true hyperbolic identity.

#### 4. Simple Hyperbolic Identities Proofs

Identity: $$\cosh^2 x – \sinh^2 x = 1$$

Proof:
$\cosh^2 x = \left( \frac{e^x + e^{-x}}{2} \right)^2 = \frac{e^{2x} + 2 + e^{-2x}}{4}$
$\sinh^2 x = \left( \frac{e^x – e^{-x}}{2} \right)^2 = \frac{e^{2x} – 2 + e^{-2x}}{4}$
$\cosh^2 x – \sinh^2 x = \frac{e^{2x} + 2 + e^{-2x}}{4} – \frac{e^{2x} – 2 + e^{-2x}}{4} = 1$

#### 5. Equations Involving Hyperbolic Functions

Example: Solve $$\cosh x = 2$$

$\cosh x = 2$
$\frac{e^x + e^{-x}}{2} = 2$
$e^x + e^{-x} = 4$
$e^{2x} + 1 = 4e^x$
$e^{2x} – 4e^x + 1 = 0$
$(e^x – 2)^2 = 0$
$e^x = 2$
$x = \ln 2$

#### 6. Derivatives and Series Expansions for Hyperbolic Functions

Derivatives:

– $$\frac{d}{dx} \cosh x = \sinh x$$
– $$\frac{d}{dx} \sinh x = \cosh x$$

#### Series Expansions:

Hyperbolic Cosine:
$\cosh x = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}$

Hyperbolic Sine:
$\sinh x = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$

Examples:

– Find the first three terms in the series expansion of $$\cosh x$$:

$\cosh x = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots$

For $$x = 1$$:

$\cosh 1 \approx 1 + \frac{1^2}{2} + \frac{1^4}{24} = 1 + 0.5 + 0.04167 \approx 1.54167$

– Find the first three terms in the series expansion of $$\sinh x$$:

$\sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots$

For $$x = 1$$:

$\sinh 1 \approx 1 + \frac{1^3}{6} + \frac{1^5}{120} = 1 + 0.16667 + 0.00833 \approx 1.175$