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9231 · 2.1

Hyperbolic functions — FAQ

Frequently asked questions for 9231 Hyperbolic functions. Direct answers first, then deeper explanation — then practise with marking.

Why are they called 'hyperbolic' functions?

They are called hyperbolic because they parametrise the unit hyperbola x2y2=1x^2 - y^2 = 1 with the point (cosh(t),sinh(t))(\cosh(t), \sinh(t)), in the same way that trigonometric functions parametrise the unit circle x2+y2=1x^2 + y^2 = 1 with the point (cos(t),sin(t))(\cos(t), \sin(t)).

What is the difference between the graph of y = cosh(x) and a parabola like y = x² + 1?

Although they look similar near the origin, they are different curves. The catenary, y=cosh(x)y = \cosh(x), grows exponentially, so it becomes much steeper than the parabola y=x2+1y = x^2 + 1 as xx moves away from zero. A parabola is a quadratic function, while a catenary is based on exponential functions.

Do I need to memorise the proofs of the identities?

Yes, it is highly recommended. Exam questions can explicitly ask you to prove an identity from the definitions, as shown in the worked example. Understanding the proofs also deepens your grasp of the relationship between hyperbolic and exponential functions.

When solving an equation, how do I choose between using definitions and using identities?

If the equation contains a mix of different hyperbolic functions (e.g., sinh(x)\sinh(x) and cosh(x)\cosh(x)), using the exponential definitions is often the most direct route. If the equation involves powers of a single function (e.g., cosh2(x)\cosh^2(x)) and another function, try using an identity like sinh2(x)=cosh2(x)1\sinh^2(x) = \cosh^2(x) - 1 to create a polynomial in one hyperbolic function.