| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Prove identity with double/compound angles |
| Difficulty | Standard +0.3 This is a structured multi-part question with clear scaffolding. Part (a) requires basic coordinate geometry and angle reasoning, part (b) applies standard double angle formulae (given the topic), part (c) is straightforward optimization by inspection, and part (d) uses Pythagoras. While it has multiple steps, each individual step is routine and the question guides students through the process with minimal problem-solving insight required. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x=d\cos\theta - d\cos2\theta = d\cos\theta - d\sin\!\left(\frac{\pi}{2}-2\theta\right) = d\!\left(\cos\theta+\sin\!\left(2\theta-\frac{\pi}{2}\right)\right)\) | M1 (3.3) | Begins to find required horizontal distance by considering appropriate horizontal distances in right-angled triangle e.g. \(d\cos\theta\), \(d\cos2\theta\) or \(d\sin\!\left(\frac{\pi}{2}-2\theta\right)\); award for correctly identifying at least one horizontal component |
| Completes correct manipulation to show required result | R1 (2.1) | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x=d(\cos\theta - \cos2\theta) = d(\cos\theta-(2\cos^2\theta-1))\) | M1 (3.1a) | Uses compound angle formula and expands \(\sin\!\left(2\theta-\frac{\pi}{2}\right)\); or uses complementary angles to obtain formula without \(\frac{\pi}{2}\); or states \(x=d(\cos\theta-\cos2\theta)\) |
| \(x=d(1+\cos\theta-2\cos^2\theta)\) | R1 (2.1) | Uses \(\cos2\theta=2\cos^2\theta-1\) to show required result; AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Greatest value \(=\frac{9d}{8}\) | B1 (1.1b) | |
| \(\cos\theta=\frac{1}{4}\) | B1 (1.1b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(OQ^2 = d^2 + d^2 - 2d^2\cos\theta\) | M1 (AO3.1a) | Begins to find OQ by cosine rule or sine rule with \(d\) and \(\theta\); accept either form |
| \(= 2d^2 - 2d^2 \times \frac{1}{4} = \frac{3d^2}{2}\) | M1 (AO1.1a) | Substitutes exact value for \(\cos\theta\) into cosine rule; note \(\cos\theta = \frac{1}{4}\), \(\sin\theta = \frac{\sqrt{15}}{4}\), \(\cos\frac{\theta}{2} = \sqrt{\frac{5}{8}}\) |
| \(OQ = \frac{\sqrt{6}}{2}d\) | A1 (AO1.1b) | Correct exact value of \(OQ\); ACF |
## Question 9(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x=d\cos\theta - d\cos2\theta = d\cos\theta - d\sin\!\left(\frac{\pi}{2}-2\theta\right) = d\!\left(\cos\theta+\sin\!\left(2\theta-\frac{\pi}{2}\right)\right)$ | M1 (3.3) | Begins to find required horizontal distance by considering appropriate horizontal distances in right-angled triangle e.g. $d\cos\theta$, $d\cos2\theta$ or $d\sin\!\left(\frac{\pi}{2}-2\theta\right)$; award for correctly identifying at least one horizontal component |
| Completes correct manipulation to show required result | R1 (2.1) | AG |
## Question 9(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x=d(\cos\theta - \cos2\theta) = d(\cos\theta-(2\cos^2\theta-1))$ | M1 (3.1a) | Uses compound angle formula and expands $\sin\!\left(2\theta-\frac{\pi}{2}\right)$; or uses complementary angles to obtain formula without $\frac{\pi}{2}$; or states $x=d(\cos\theta-\cos2\theta)$ |
| $x=d(1+\cos\theta-2\cos^2\theta)$ | R1 (2.1) | Uses $\cos2\theta=2\cos^2\theta-1$ to show required result; AG |
## Question 9(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Greatest value $=\frac{9d}{8}$ | B1 (1.1b) | |
| $\cos\theta=\frac{1}{4}$ | B1 (1.1b) | |
## Question 9(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $OQ^2 = d^2 + d^2 - 2d^2\cos\theta$ | M1 (AO3.1a) | Begins to find OQ by cosine rule or sine rule with $d$ and $\theta$; accept either form |
| $= 2d^2 - 2d^2 \times \frac{1}{4} = \frac{3d^2}{2}$ | M1 (AO1.1a) | Substitutes exact value for $\cos\theta$ into cosine rule; note $\cos\theta = \frac{1}{4}$, $\sin\theta = \frac{\sqrt{15}}{4}$, $\cos\frac{\theta}{2} = \sqrt{\frac{5}{8}}$ |
| $OQ = \frac{\sqrt{6}}{2}d$ | A1 (AO1.1b) | Correct exact value of $OQ$; ACF |
---9 A robotic arm which is attached to a flat surface at the origin $O$, is used to draw a graphic design.
The arm is made from two rods $O P$ and $P Q$, each of length $d$, which are joined at $P$.\\
A pen is attached to the arm at $Q$.\\
The coordinates of the pen are controlled by adjusting the angle $O P Q$ and the angle $\theta$ between $O P$ and the $x$-axis.
For this particular design the pen is made to move so that the two angles are always equal to each other with $0 \leq \theta \leq \frac { \pi } { 2 }$ as shown in Figure 2.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{b7df05bf-f3fc-4705-a13c-6b562896fa9f-12_805_867_989_584}
\end{center}
\end{figure}
9
\begin{enumerate}[label=(\alph*)]
\item Show that the $x$-coordinate of the pen can be modelled by the equation
$$x = d \left( \cos \theta + \sin \left( 2 \theta - \frac { \pi } { 2 } \right) \right)$$
9
\item Hence, show that
$$x = d \left( 1 + \cos \theta - 2 \cos ^ { 2 } \theta \right)$$
9
\item It can be shown that
$$x = \frac { 9 d } { 8 } - d \left( \cos \theta - \frac { 1 } { 4 } \right) ^ { 2 }$$
State the greatest possible value of $x$ and the corresponding value of $\cos \theta$\\
9
\item Figure 3 below shows the arm when the $x$-coordinate is at its greatest possible value.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\includegraphics[alt={},max width=\textwidth]{b7df05bf-f3fc-4705-a13c-6b562896fa9f-14_570_773_456_630}
\end{center}
\end{figure}
Find, in terms of $d$, the exact distance $O Q$.\\
\includegraphics[max width=\textwidth, alt={}, center]{b7df05bf-f3fc-4705-a13c-6b562896fa9f-15_2488_1716_219_153}
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2021 Q9 [9]}}