| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2008 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Given sin/cos/tan, find other expressions |
| Difficulty | Moderate -0.3 This is a straightforward application of standard trigonometric identities with no novel problem-solving required. Part (a) uses Pythagorean identity and compound angle formula with given values; part (b) applies the tan(2x) double angle formula and solves a quadratic. All techniques are routine for C4 students, making it slightly easier than average but not trivial due to the multi-step nature and algebraic manipulation required. |
| Spec | 1.05g Exact trigonometric values: for standard angles1.05l Double angle formulae: and compound angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos\alpha = \frac{3}{5}\) | B1 | ACF |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos(\alpha-\beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta\) | M1 | |
| \(= \frac{3}{5}\cos\beta + \frac{4}{5}\sin\beta\) | A1 | ACF |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sin\beta = \frac{12}{13}\) | B1 | |
| \(\cos(\alpha-\beta) = \frac{63}{65}\) | B1 | \(\frac{63}{65}\) NMS B1B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan 2x = \frac{2\tan x}{1-\tan^2 x}\) | M1 | |
| \(2\tan x = 1 - \tan^2 x\) leading to \(\tan^2 x + 2\tan x - 1 = 0\) | A1 | CSO AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan x = \frac{-2\pm\sqrt{4+4}}{2}\) | M1 | Must solve quadratic by formula or completing the square; condone one slip |
| \(= -1\pm\sqrt{2}\) | A1 | \(\pm\sqrt{2}\) required |
| \(2x = 45°\Rightarrow x = 22\frac{1}{2}°\) is acute | ||
| \(\Rightarrow \tan 22\frac{1}{2}° = \sqrt{2}-1\) | E1 | Explain selection of positive root |
# Question 5:
## Part 5(a)(i)
$\cos\alpha = \frac{3}{5}$ | B1 | ACF
## Part 5(a)(ii)
$\cos(\alpha-\beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta$ | M1 |
$= \frac{3}{5}\cos\beta + \frac{4}{5}\sin\beta$ | A1 | ACF
## Part 5(a)(iii)
$\sin\beta = \frac{12}{13}$ | B1 |
$\cos(\alpha-\beta) = \frac{63}{65}$ | B1 | $\frac{63}{65}$ NMS B1B1
## Part 5(b)(i)
$\tan 2x = \frac{2\tan x}{1-\tan^2 x}$ | M1 |
$2\tan x = 1 - \tan^2 x$ leading to $\tan^2 x + 2\tan x - 1 = 0$ | A1 | CSO AG
## Part 5(b)(ii)
$\tan x = \frac{-2\pm\sqrt{4+4}}{2}$ | M1 | Must solve quadratic by formula or completing the square; condone one slip
$= -1\pm\sqrt{2}$ | A1 | $\pm\sqrt{2}$ required
$2x = 45°\Rightarrow x = 22\frac{1}{2}°$ is acute | |
$\Rightarrow \tan 22\frac{1}{2}° = \sqrt{2}-1$ | E1 | Explain selection of positive root
---5
\begin{enumerate}[label=(\alph*)]
\item The angle $\alpha$ is acute and $\sin \alpha = \frac { 4 } { 5 }$.
\begin{enumerate}[label=(\roman*)]
\item Find the value of $\cos \alpha$.
\item Express $\cos ( \alpha - \beta )$ in terms of $\sin \beta$ and $\cos \beta$.
\item Given also that the angle $\beta$ is acute and $\cos \beta = \frac { 5 } { 13 }$, find the exact value of $\cos ( \alpha - \beta )$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Given that $\tan 2 x = 1$, show that $\tan ^ { 2 } x + 2 \tan x - 1 = 0$.
\item Hence, given that $\tan 45 ^ { \circ } = 1$, show that $\tan 22 \frac { 1 } { 2 } ^ { \circ } = \sqrt { 2 } - 1$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C4 2008 Q5 [10]}}