AS June 2021 Paper 1 Q6

AQACurrent spec2 marksHyperbolic Functions

6 Prove the identity

\[\cosh^2 x - \sinh^2 x = 1\]

[2 marks]

Mark scheme

Scheme	Marks	AO
Recalls exponential definitions of hyperbolic functions and substitutes \(\dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\) and \(\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\) Condone \(\cosh x\) replaced with \(\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\) and \(\sinh x\) replaced with \(\dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\)	M1	1.2
Completes a fully correct proof to reach the required result.	R1	2.1
	(2 marks)

Typical solution

\[\begin{aligned}\cosh^2 x - \sinh^2 x &= \left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^2 - \left(\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right)^2 \\ &= \frac{\mathrm{e}^{2x} + 2 + \mathrm{e}^{-2x}}{4} - \frac{\mathrm{e}^{2x} - 2 + \mathrm{e}^{-2x}}{4} \\ &= \frac{4}{4} \\ &= 1\end{aligned}\]