Edexcel A-Level Mathematics A2 Fp3 1 Hyperbolic Functions Questions

Hyperbolic-function questions are proof-led: expand definitions in exponentials, match coefficients in R sinh forms and finish with exact logarithms.

Exam points

Replace sinh, cosh, tanh or sech with exponentials to prove identities.
Match Rcoshα and Rsinhα when rewriting a sinh x+b cosh x as R sinh(x+α).
Solve hyperbolic equations and state exact answers as simplified logarithms.

Covered syllabus points

FP3.1.1 - Definition of the six hyperbolic FP3.1.2 - Inverse hyperbolic functions

[Maximum number: 9]

(a)

Use the definitions of hyperbolic functions in terms of exponentials to show that

\sinh(A+B) \equiv \sinh A \cosh B + \cosh A \sinh B

[ 3 ]

\sinh A\cosh B+\cosh A\sinh B =\frac{e^A-e^{-A}}{2}\frac{e^B+e^{-B}}{2} +\frac{e^A+e^{-A}}{2}\frac{e^B-e^{-B}}{2}

=\frac{2e^{A+B}-2e^{-(A+B)}}{4}=\sinh(A+B)

M1: replaces two of the four hyperbolic functions with correct exponential expressions.
M1: expands the numerator with at least four terms.
A1*: reaches sinh(A+B) with no errors.
There is no credit for proofs that do not use exponential definitions.

(b)

Hence express $10\sinh x+8\cosh x$ in the form $R\sinh(x+\alpha)$ where R>0,
giving $\alpha$ in the form $\ln p$ where p is an integer.

[ 4 ]

R\sinh(x+\alpha)=R\cosh\alpha\sinh x+R\sinh\alpha\cosh x

R\cosh\alpha=10,\quad R\sinh\alpha=8

R^2=10^2-8^2=36\Rightarrow R=6

\tanh\alpha=\frac{8}{10}=\frac45

\alpha=\operatorname{artanh}\left(\frac45\right)=\frac12\ln\left(\frac{1+4/5}{1-4/5}\right)=\ln 3

10\sinh x+8\cosh x=6\sinh(x+\ln 3)

B1/M1: equates coefficients correctly.
M1: complete attempt to find positive R.
M1: complete attempt to find alpha = ln p.
A1: R = 6, alpha = ln 3, so the expression is 6sinh(x + ln 3).

(c)

Hence solve the equation

10\sinh x+8\cosh x=18\sqrt7

giving your answer in the form $\ln(\sqrt7+q)$ where q is a rational number to be determined.

[ 2 ]

Using part (b),

6\sinh(x+\ln 3)=18\sqrt{7}

\sinh(x+\ln 3)=3\sqrt{7}

x+\ln 3=\operatorname{arsinh}(3\sqrt{7})

x=\ln\left(3\sqrt{7}+\sqrt{(3\sqrt{7})^2+1}\right)-\ln 3

x=\ln\left(3\sqrt{7}+8\right)-\ln 3=\ln\left(\sqrt{7}+\frac{8}{3}\right)

M1: uses part (b) to form the equation in hyperbolic sine.
M1: applies the inverse hyperbolic sine form.
A1: $x=\ln\left(\sqrt{7}+\frac{8}{3}\right)$ , so $q=\frac{8}{3}$ .

[Maximum number: 6]

Given that

\cosh y=x \quad \text { and } \quad y<0

use the definition of $\cosh y$ in terms of exponential functions to prove that

y=\ln \left(x-\sqrt{x^{2}-1}\right)

$\cosh y=x,\quad y<0,\quad x=\frac{e^y+e^{-y}}{2}$ $2xe^y=e^{2y}+1 \Rightarrow e^{2y}-2xe^y+1=0$ $e^y=\frac{2x\pm\sqrt{(2x)^2-4}}{2} =x\pm\sqrt{x^2-1}$ $y=\ln\left(x-\sqrt{x^2-1}\right)$

Since y<0, $e^y<1$ , so the negative root is required.

B1 M1 M1 A1 A1* B1

[Maximum number: 10]

(a)

Use the definitions of $\sinh x$ and $\cosh x$ in terms of exponentials to show that

\cosh A\cosh B+\sinh A\sinh B\equiv \cosh(A+B)

[ 5 ]

$\cosh A\cosh B+\sinh A\sinh B =\left(\frac{e^A+e^{-A}}{2}\right)\left(\frac{e^B+e^{-B}}{2}\right) +\left(\frac{e^A-e^{-A}}{2}\right)\left(\frac{e^B-e^{-B}}{2}\right)$ $=\frac{e^{A+B}+e^{A-B}+e^{B-A}+e^{-A-B}+e^{A+B}-e^{A-B}-e^{B-A}+e^{-A-B}}{4}$ $=\frac{2e^{A+B}+2e^{-(A+B)}}{4} =\frac{e^{A+B}+e^{-(A+B)}}{2} =\cosh(A+B).$

M1 A1*

(b)

Hence find the value of x for which

\cosh(x+\ln2)=5\sinh x

giving your answer in the form $\frac12\ln k$ , where k is a rational number to be determined.

[ 5 ]

\cosh(x+\ln2)=\cosh x\cosh(\ln2)+\sinh x\sinh(\ln2)

=\frac54\cosh x+\frac34\sinh x

\frac54\cosh x+\frac34\sinh x=5\sinh x

5\cosh x=17\sinh x

\tanh x=\frac5{17}

x=\frac12\ln\left(\frac{1+\frac5{17}}{1-\frac5{17}}\right)=\frac12\ln\left(\frac{11}{6}\right)

All question bank results loaded

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question 2

Question 3

Question (a)

Question (b)