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What are the hyperbolic functions, what identities do they satisfy, and what do their inverses look like?

The hyperbolic functions defined from exponentials, their graphs and properties, the key identities, and the logarithmic forms of the inverse hyperbolic functions.

A focused answer to the OCR A-Level Further Mathematics A content on hyperbolic functions, covering the definitions of sinh, cosh and tanh from exponentials, their graphs and odd or even properties, the key identities such as cosh squared minus sinh squared equals one, and the logarithmic forms of the inverse hyperbolic functions.

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  1. What this dot point is asking
  2. Definitions from exponentials
  3. Graphs and properties
  4. The key identities
  5. The inverse hyperbolic functions
  6. Try this

What this dot point is asking

OCR wants you to define the hyperbolic functions sinh\sinh, cosh\cosh and tanh\tanh in terms of exponentials, sketch and describe their graphs (including odd or even symmetry and asymptotes), use the key identities (above all cosh2xsinh2x=1\cosh^2 x - \sinh^2 x = 1), and derive and use the logarithmic forms of the inverse hyperbolic functions.

Definitions from exponentials

The hyperbolic functions are combinations of exe^x and exe^{-x}, named by analogy with the circular (trigonometric) functions. They appear naturally in calculus, in the catenary (a hanging chain) and in solutions of differential equations.

Graphs and properties

The graphs follow from the definitions. coshx\cosh x is even (symmetric about the yy-axis), always 1\ge 1, with a minimum at (0,1)(0, 1) and both arms rising like 12ex\tfrac{1}{2}e^{|x|}. sinhx\sinh x is odd, passing through the origin and increasing throughout. tanhx\tanh x is odd, increasing, sandwiched between the horizontal asymptotes y=1y = 1 and y=1y = -1.

The key identities

The hyperbolic identities mirror the trigonometric ones, with sign changes captured by Osborn's rule. The central one follows directly from the definitions.

The inverse hyperbolic functions

Because the hyperbolic functions are built from exponentials, their inverses are logarithms. You derive each by setting yy equal to the inverse, writing the defining exponential equation, and solving the resulting quadratic in eye^y. The same three-step pattern works every time: write x=sinhyx = \sinh y (or coshy\cosh y, or tanhy\tanh y) in exponential form, multiply through by a power of eye^y to obtain a quadratic in eye^y, then solve and take logarithms, rejecting any root that would make eye^y negative. Memorising the final logarithmic forms is useful for speed, but you must be able to derive them on demand, because OCR sometimes asks for the derivation as a "show that" question worth several marks.

Hyperbolic functions feed directly into the calculus topic, where their derivatives and integrals (and the inverse forms as standard integrals) are examined.

Try this

Q1. State the value of sinh0\sinh 0 and cosh0\cosh 0. [2 marks]

  • Cue. sinh0=112=0\sinh 0 = \dfrac{1 - 1}{2} = 0; cosh0=1+12=1\cosh 0 = \dfrac{1 + 1}{2} = 1.

Q2. Use the identity to find cosh2x\cosh^2 x if sinhx=34\sinh x = \tfrac{3}{4}. [2 marks]

  • Cue. cosh2x=1+sinh2x=1+916=2516\cosh^2 x = 1 + \sinh^2 x = 1 + \tfrac{9}{16} = \tfrac{25}{16}.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20195 marksStarting from the definitions of sinhx\sinh x and coshx\cosh x, prove that cosh2xsinh2x=1\cosh^2 x - \sinh^2 x = 1.
Show worked answer →

Use the definitions (M1): coshx=ex+ex2\cosh x = \dfrac{e^x + e^{-x}}{2}, sinhx=exex2\sinh x = \dfrac{e^x - e^{-x}}{2}.

Square each (M1): cosh2x=e2x+2+e2x4\cosh^2 x = \dfrac{e^{2x} + 2 + e^{-2x}}{4}, sinh2x=e2x2+e2x4\sinh^2 x = \dfrac{e^{2x} - 2 + e^{-2x}}{4} (A1).

Subtract (A1): cosh2xsinh2x=(e2x+2+e2x)(e2x2+e2x)4=44=1\cosh^2 x - \sinh^2 x = \dfrac{(e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x})}{4} = \dfrac{4}{4} = 1 (A1).

Markers reward the definitions, squaring both, subtracting, and simplifying to 11.

OCR 20225 marksShow that arsinhx=ln(x+x2+1)\text{arsinh}\,x = \ln\left(x + \sqrt{x^2 + 1}\right).
Show worked answer →

Let y=arsinhxy = \text{arsinh}\,x, so x=sinhy=eyey2x = \sinh y = \dfrac{e^y - e^{-y}}{2} (M1).

Multiply by 2ey2e^y to clear (M1): 2xey=e2y12x e^y = e^{2y} - 1, so e2y2xey1=0e^{2y} - 2x e^y - 1 = 0, a quadratic in eye^y (A1).

Solve for eye^y by the quadratic formula: ey=2x±4x2+42=x±x2+1e^y = \dfrac{2x \pm \sqrt{4x^2 + 4}}{2} = x \pm \sqrt{x^2 + 1} (M1). Since ey>0e^y > 0, take the ++ sign: ey=x+x2+1e^y = x + \sqrt{x^2 + 1}.

Take logs (A1): y=ln(x+x2+1)y = \ln\left(x + \sqrt{x^2 + 1}\right).

Markers reward setting x=sinhyx = \sinh y, forming the quadratic in eye^y, solving and rejecting the negative root, and taking logarithms.

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