Definitions of sinh, cosh and tanh from exponentials, hyperbolic identities, logarithmic forms of the inverse functions, and differentiation and integration of hyperbolic functions.

What this dot point is asking

Edexcel wants you to define $\sinh$, $\cosh$ and $\tanh$ from the exponential function, use the hyperbolic identities (including via Osborn's rule), write the inverse functions in logarithmic form and derive those forms, and differentiate and integrate hyperbolic functions, including the standard integrals they produce. Hyperbolic functions are a Core Pure staple and feed directly into the further-calculus standard integrals.

Definitions and identities

Hyperbolic functions are built directly from $e^x$ and $e^{-x}$. From the definitions, $\cosh$ is even (symmetric about the $y$-axis, with minimum value $1$ at $x = 0$) and $\sinh$ is odd (rotationally symmetric about the origin, passing through it). As $x \to \infty$ both behave like $\frac{1}{2}e^x$, and $\tanh x \to 1$.

Osborn's rule converts a familiar trigonometric identity to its hyperbolic counterpart: replace $\cos$ by $\cosh$ and $\sin$ by $\sinh$, but change the sign of any term containing a product of two sines (because $\sin^2$ implicitly carries an $i^2 = -1$ when you compare with the complex link). For instance $\cos^2\theta + \sin^2\theta = 1$ becomes $\cosh^2 x - \sinh^2 x = 1$.

Inverse hyperbolic functions

Because the definitions are exponential, the inverses are logarithmic, and you derive them by solving a quadratic in $e^y$. These logarithmic forms are exactly what you quote when an integral evaluates to an inverse hyperbolic function but the question asks for an exact logarithm.

Differentiation and integration

The derivatives match the trigonometric pattern but with no sign change between $\sinh$ and $\cosh$, and reversing them produces the standard inverse-hyperbolic integrals used throughout further calculus.

Examples in context

Hyperbolic functions thread through Further Maths. Their inverses give the standard integrals $\int \frac{1}{\sqrt{x^2 \pm a^2}}\,dx$ in the further-calculus dot point, and the logarithmic forms convert those answers into exact natural logs. The link to complex numbers is direct: $\cos(ix) = \cosh x$ and $\sin(ix) = i\sinh x$, which is the deep reason Osborn's rule works. Hyperbolic substitution ($x = a\sinh u$ or $x = a\cosh u$) is the natural method for integrals containing $\sqrt{x^2 + a^2}$ or $\sqrt{x^2 - a^2}$. The catenary, the curve of a hanging chain, is $y = a\cosh\frac{x}{a}$, a classic modelling appearance, and $\cosh$ also arises as the complementary function of differential equations with real auxiliary roots $\pm m$.

Try this

Q1. Show that $\cosh^2 x - \sinh^2 x = 1$ from the definitions. [3 marks]

• Cue. Square each, subtract, and the cross terms cancel leaving $1$.

Q2. Differentiate $y = \cosh 3x$. [2 marks]

• Cue. Chain rule: $\frac{dy}{dx} = 3\sinh 3x$.

Q3. Solve $\cosh x = 2$, giving exact logarithmic answers. [3 marks]

• Cue. $\operatorname{arcosh} 2 = \ln(2 + \sqrt{3})$, with $x = \pm\ln(2 + \sqrt{3})$ since $\cosh$ is even.