A2 June 2020 Paper 1 Q14

AQACurrent spec6 marksHyperbolic Functions Series

(a) Given that\[\sinh(A + B) = \sinh A\cosh B + \cosh A\sinh B\]

express \(\sinh(m + 1)x\) and \(\sinh(m - 1)x\) in terms of \(\sinh mx\), \(\cosh mx\), \(\sinh x\) and \(\cosh x\) [1 mark]

(b) Hence find the sum of the series\[C_n = \cosh x + \cosh 2x + \cdots + \cosh nx\]

in terms of \(\sinh x\), \(\sinh nx\) and \(\sinh(n + 1)x\) [5 marks]

Mark scheme (a)

Scheme	Marks	AO
Writes the correct expressions for \(\sinh(m + 1)x\) and \(\sinh(m - 1)x\)	B1	1.1b

Typical solution

\[\sinh(m + 1)x = \sinh mx\cosh x + \cosh mx\sinh x\]\[\sinh(m - 1)x = \sinh mx\cosh x - \cosh mx\sinh x\]

Mark scheme (b)

Scheme	Marks	AO
Subtracts \(\sinh(m + 1)x\) and \(\sinh(m - 1)x\) expressions.	M1	3.1a
Uses the method of differences, with at least the first two terms shown.	M1	3.1a
Obtains correct terms for method of differences, with at least the first three terms and the last two.	A1	1.1b
Deduces that \(\sum 2\cosh mx\sinh x\) is a multiple of \(C_n\)	M1	2.2a
Completes a rigorous argument to show the required result.	R1	2.1
	(6 marks)

Typical solution

\[\sinh(m + 1)x - \sinh(m - 1)x = 2\cosh mx\sinh x\]\[\begin{aligned} \cancel{\sinh 2x} - \sinh 0 &= 2\cosh x\sinh x \\ \cancel{\sinh 3x} - \sinh x &= 2\cosh 2x\sinh x \\ \cancel{\sinh 4x} - \cancel{\sinh 2x} &= 2\cosh 3x\sinh x \\ &\ \,\vdots \\ \cancel{\sinh(n - 1)x} - \cancel{\sinh(n - 3)x} &= 2\cosh(n - 2)x\sinh x \\ \sinh nx - \cancel{\sinh(n - 2)x} &= 2\cosh(n - 1)x\sinh x \\ \sinh(n + 1)x - \cancel{\sinh(n - 1)x} &= 2\cosh nx\sinh x \end{aligned}\]\[\begin{aligned} \sinh(n + 1)x + \sinh nx - \sinh x &= 2\sinh x(\cosh x + \cosh 2x + \cdots + \cosh nx) \\ &= 2\sinh x\,C_n \end{aligned}\]\[\therefore C_n = \frac{\sinh(n + 1)x + \sinh nx - \sinh x}{2\sinh x}\]