| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2016 |
| 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 | Standard +0.3 This is a standard C4 trigonometric identities question requiring routine application of Pythagorean identity, double angle formula, and compound angle formulas. While it involves multiple steps and algebraic manipulation with surds, all techniques are textbook exercises with no novel insight required. The 'show that' format provides target answers, making it slightly easier than average. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Use of \(\sin^2B + \cos^2B = 1\) | ||
| \(\left(\frac{1}{\sqrt{5}}\right)^2 + \cos^2B = 1\) | ||
| \(\cos B = \frac{2}{\sqrt{5}}\) | B1 | Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \((\sin 2B = 2\sin B \cos B)\) | ||
| \(= 2 \times \frac{1}{\sqrt{5}} \times \frac{2}{\sqrt{5}}\) | M1 | Correct identity (PI) and substitution |
| \(= \frac{4}{5}\) | A1 | Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos A = \frac{2}{3}\) exact value | B1 | \(\cos A = \frac{2}{3}\) seen or used (not 0.667 etc.) |
| \(\sin(A - B) = \sin A \cos B - \cos A \sin B\) | ||
| \(= \frac{\sqrt{5}}{3} \times \frac{2}{\sqrt{5}} - \frac{2}{3} \times \frac{1}{\sqrt{5}}\) | M1 | ft on their value of cos A |
| Use of \(\frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}\) or \(\frac{10 - 2\sqrt{5}}{15}\) OE seen | m1 | \(\frac{2}{3\sqrt{5}}\) term becoming \(\frac{2\sqrt{5}}{15}\) before final answer |
| \(\frac{2}{15}(5 - \sqrt{5})\) | A1 | Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos(A - B) = \cos A \cos B + \sin A \sin B\) | ||
| \(= \frac{2}{3} \times \frac{2}{\sqrt{5}} + \frac{\sqrt{5}}{3} \times \frac{1}{\sqrt{5}}\) | M1 | ft on their value of \(\cos A\) |
| A1 | fully correct | |
| \(= \frac{1}{3} + \frac{4}{15}\sqrt{5}\) | A1 | Total: 3 |
| Total | 10 |
## Part (a)(i)
Use of $\sin^2B + \cos^2B = 1$ | |
$\left(\frac{1}{\sqrt{5}}\right)^2 + \cos^2B = 1$ | |
$\cos B = \frac{2}{\sqrt{5}}$ | B1 | Total: 1 | $\cos(\sin^{-1}(\frac{1}{\sqrt{5}})) = \frac{2}{\sqrt{5}}$ is B0
AG ; must see evidence of working
Or use of right-angled triangle with opp = 1 and hyp = $\sqrt{5}$ to get adj = $\sqrt{4}$ or 2.
## Part (a)(ii)
$(\sin 2B = 2\sin B \cos B)$ | |
$= 2 \times \frac{1}{\sqrt{5}} \times \frac{2}{\sqrt{5}}$ | M1 | Correct identity (PI) and substitution
$= \frac{4}{5}$ | A1 | Total: 2 | AG so line above must be seen.
## Part (b)(i)
$\cos A = \frac{2}{3}$ exact value | B1 | $\cos A = \frac{2}{3}$ seen or used (not 0.667 etc.)
$\sin(A - B) = \sin A \cos B - \cos A \sin B$ | |
$= \frac{\sqrt{5}}{3} \times \frac{2}{\sqrt{5}} - \frac{2}{3} \times \frac{1}{\sqrt{5}}$ | M1 | ft on their value of cos A
Use of $\frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}$ or $\frac{10 - 2\sqrt{5}}{15}$ OE seen | m1 | $\frac{2}{3\sqrt{5}}$ term becoming $\frac{2\sqrt{5}}{15}$ before final answer
$\frac{2}{15}(5 - \sqrt{5})$ | A1 | Total: 4 | $\frac{2}{15}$ OE seen and be convinced
You must see justification between the use of the identity and the final answer to earn the m1 A1.
## Part (b)(ii)
$\cos(A - B) = \cos A \cos B + \sin A \sin B$ | |
$= \frac{2}{3} \times \frac{2}{\sqrt{5}} + \frac{\sqrt{5}}{3} \times \frac{1}{\sqrt{5}}$ | M1 | ft on their value of $\cos A$
| A1 | fully correct
$= \frac{1}{3} + \frac{4}{15}\sqrt{5}$ | A1 | Total: 3 | OE for $\frac{1}{3}$ and $\frac{4}{15}$ but not left as $\frac{5+4\sqrt{5}}{15}$
Total | | 10
---It is given that $\sin A = \frac{\sqrt{5}}{3}$ and $\sin B = \frac{1}{\sqrt{5}}$, where the angles $A$ and $B$ are both acute.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the exact value of $\cos B = \frac{2}{\sqrt{5}}$.
[1 mark]
\item Hence show that the exact value of $\sin 2B$ is $\frac{4}{5}$.
[2 marks]
\end{enumerate}
\item \begin{enumerate}[label=(\roman*)]
\item Show that the exact value of $\sin(A - B)$ can be written as $p(5 - \sqrt{5})$, where $p$ is a rational number.
[4 marks]
\item Find the exact value of $\cos(A - B)$ in the form $r + s\sqrt{5}$, where $r$ and $s$ are rational numbers.
[3 marks]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2016 Q5 [10]}}