AQA FP2 2011 June — Question 2 10 marks

Exam BoardAQA
ModuleFP2 (Further Pure Mathematics 2)
Year2011
SessionJune
Marks10
PaperDownload PDF ↗
TopicHyperbolic functions
TypeProve hyperbolic identity from exponentials
DifficultyStandard +0.3 This is a standard FP2 hyperbolic functions question requiring routine manipulation of definitions and identities. Part (a) is direct substitution of exponential definitions (textbook exercise), part (b)(i) involves applying the result from (a) and simplifying, and part (b)(ii) is straightforward application of the inverse tanh formula. While it requires multiple steps, all techniques are standard for FP2 with no novel insight needed.
Spec4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 1

  1. Use the definitions of \(\cosh \theta\) and \(\sinh \theta\) in terms of \(e^\theta\) to show that $$\cosh x \cosh y - \sinh x \sinh y = \cosh(x - y)$$ [4 marks]
  2. It is given that \(x\) satisfies the equation $$\cosh(x - \ln 2) = \sinh x$$
    1. Show that \(\tanh x = \frac{5}{4}\). [4 marks]
    2. Express \(x\) in the form \(\frac{1}{2} \ln a\). [2 marks]

\begin{enumerate}[label=(\alph*)]
\item Use the definitions of $\cosh \theta$ and $\sinh \theta$ in terms of $e^\theta$ to show that
$$\cosh x \cosh y - \sinh x \sinh y = \cosh(x - y)$$ [4 marks]

\item It is given that $x$ satisfies the equation
$$\cosh(x - \ln 2) = \sinh x$$
\begin{enumerate}[label=(\roman*)]
\item Show that $\tanh x = \frac{5}{4}$. [4 marks]
\item Express $x$ in the form $\frac{1}{2} \ln a$. [2 marks]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{AQA FP2 2011 Q2 [10]}}