| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| 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 proof-completion question with scaffolding at each step. Parts (a)-(c) involve explaining a geometric proof with most work done, requiring only identification of ratios and understanding of the diagram. Part (d) is a standard application of odd/even function properties. While it tests understanding of the addition formula derivation, the heavy scaffolding and routine nature of part (d) make this easier than average. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.05m Geometric proofs: of trig sum and double angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| \(\angle OQR = \angle FQE\) vertically opposite angles | E1 | Must be fully correct explanation with reasons |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{PF}{EF} = \cos(A)\) | R1 | Must have stated or implied \(\angle EFQ = A\) through similarity |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{DE}{OE} \times \frac{OE}{OF} + \frac{PF}{EF} \times \frac{EF}{OF}\) | B1 | Completes proof |
| Answer | Marks | Guidance |
|---|---|---|
| Since the proof is based on the diagram which uses right-angled triangles it is assumed that \(A\) and \(B\) are acute. Therefore, the proof only holds for acute angles. | E1 | Explains proof based on right angled triangles which limits \(A\) and \(B\) to acute angles |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sin(A-B) = \sin A\cos(-B) + \cos A\sin(-B)\) | R1 | Substitutes \(-B\) into identity for \(\sin(A+B)\) to give \(\sin(A-B)\) |
| \(\sin(-B) = -\sin(B)\) | B1 | Recalls at least one identity; must be explicitly stated |
| Answer | Marks | Guidance |
|---|---|---|
| Hence \(\sin(A-B) = \sin A\cos B - \cos A\sin B\) | R1 | Deduces correct identity with no errors; must be clearly deduced from correct argument |
## Question 14(a):
$\angle OQR = \angle FQE$ vertically opposite angles | E1 | Must be fully correct explanation with reasons
$\angle ORQ = \angle FEQ = 90°$
So $\angle EFQ = A$
Since $\angle EFQ = A$:
$\frac{PF}{EF} = \cos(A)$ | R1 | Must have stated or implied $\angle EFQ = A$ through similarity
AND $\frac{EF}{OF} = \sin(B)$
---
## Question 14(b):
$\frac{DE}{OE} \times \frac{OE}{OF} + \frac{PF}{EF} \times \frac{EF}{OF}$ | B1 | Completes proof
$= \sin A \cos B + \cos A \sin B$
---
## Question 14(c):
Since the proof is based on the diagram which uses right-angled triangles it is assumed that $A$ and $B$ are acute. Therefore, the proof only holds for acute angles. | E1 | Explains proof based on right angled triangles which limits $A$ and $B$ to acute angles
---
## Question 14(d):
$\sin(A-B) = \sin A\cos(-B) + \cos A\sin(-B)$ | R1 | Substitutes $-B$ into identity for $\sin(A+B)$ to give $\sin(A-B)$
$\sin(-B) = -\sin(B)$ | B1 | Recalls at least one identity; must be explicitly stated
$\cos(-B) = \cos(B)$
Hence $\sin(A-B) = \sin A\cos B - \cos A\sin B$ | R1 | Deduces correct identity with no errors; must be clearly deduced from correct argument
**Total: 7 marks**
---14 Some students are trying to prove an identity for $\sin ( A + B )$.
They start by drawing two right-angled triangles $O D E$ and $O E F$, as shown.\\
\includegraphics[max width=\textwidth, alt={}, center]{85b10472-8149-4387-999f-2ef153f1a105-22_695_662_477_689}
The students' incomplete proof continues,\\
Let angle $D O E = A$ and angle $E O F = B$.\\
In triangle OFR,\\
Line $1 \quad \sin ( A + B ) = \frac { R F } { O F }$
Line 2
$$= \frac { R P + P F } { O F }$$
Line 3
$$= \frac { D E } { O F } + \frac { P F } { O F } \text { since } D E = R P$$
Line 4
$$= \frac { D E } { \cdots \cdots } \times \frac { \cdots \cdots } { O F } + \frac { P F } { E F } \times \frac { E F } { O F }$$
Line 5\\
$=$ $\_\_\_\_$ $+ \cos A \sin B$
14
\begin{enumerate}[label=(\alph*)]
\item Explain why $\frac { P F } { E F } \times \frac { E F } { O F }$ in Line 4 leads to $\cos A \sin B$ in Line 5\\
14
\item Complete Line 4 and Line 5 to prove the identity
Line 4
$$= \frac { D E } { \ldots \ldots } \times \frac { \cdots \ldots } { O F } + \frac { P F } { E F } \times \frac { E F } { O F }$$
Line 5 = $+ \cos A \sin B$
14
\item Explain why the argument used in part (a) only proves the identity when $A$ and $B$ are acute angles.
14
\item Another student claims that by replacing $B$ with $- B$ in the identity for $\sin ( A + B )$ it is possible to find an identity for $\sin ( A - B )$.
Assuming the identity for $\sin ( A + B )$ is correct for all values of $A$ and $B$, prove a similar result for $\sin ( A - B )$.
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2018 Q14 [7]}}