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 \( x = 1 \):
\[ \sinh 1 \approx 1 + \frac{1^3}{6} + \frac{1^5}{120} = 1 + 0.16667 + 0.00833 \approx 1.175 \]