| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Product to sum using compound angles |
| Difficulty | Moderate -0.5 This is a structured, guided question that walks students through standard compound angle formula manipulations. Parts (a) and (b) involve straightforward algebraic expansion and simplification with no problem-solving required. Part (c) applies the result mechanically with given exact values. While it requires careful algebra across multiple steps, it's easier than average because it's heavily scaffolded and uses routine techniques. |
| Spec | 1.05g Exact trigonometric values: for standard angles1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals |
| 5. continued | Leave blank |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(\sin(A+B) - \sin(A-B)\) | ||
| \(= \sin A\cos B + \sin B\cos A - \sin A\cos B + \sin B\cos A\) | M1 | |
| \(= 2\sin B\cos A\) | A1 cso | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(\cos(A-B) - \cos(A+B)\) | ||
| \(= \cos A\cos B + \sin A\sin B - \cos A\cos B + \sin A\sin B\) | M1 | |
| \(= 2\sin A\sin B\) | A1 cso | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(\frac{\sin(A+B)-\sin(A-B)}{\cos(A-B)-\cos(A+B)} = \frac{2\sin B\cos A}{2\sin A\sin B}\) | M1 | |
| \(= \frac{\cos A}{\sin A}\) | A1 | |
| \(= \cot A\) | A1 cso | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Let \(A = 75°\) and \(B = 15°\) | B1 | |
| \(\frac{\sin 90° - \sin 60°}{\cos 60° - \cos 90°} = \cot 75°\) | M1 | |
| \(\cot 75° = \frac{1 - \frac{\sqrt{3}}{2}}{\frac{1}{2} - 0} = 2 - \sqrt{3}\) | M1 A1 | (4 marks) |
# Question 5:
## Part (a)(i)
| Answer/Working | Marks | Notes |
|---|---|---|
| $\sin(A+B) - \sin(A-B)$ | | |
| $= \sin A\cos B + \sin B\cos A - \sin A\cos B + \sin B\cos A$ | M1 | |
| $= 2\sin B\cos A$ | A1 cso | (2 marks) |
## Part (a)(ii)
| Answer/Working | Marks | Notes |
|---|---|---|
| $\cos(A-B) - \cos(A+B)$ | | |
| $= \cos A\cos B + \sin A\sin B - \cos A\cos B + \sin A\sin B$ | M1 | |
| $= 2\sin A\sin B$ | A1 cso | (2 marks) |
## Part (b)
| Answer/Working | Marks | Notes |
|---|---|---|
| $\frac{\sin(A+B)-\sin(A-B)}{\cos(A-B)-\cos(A+B)} = \frac{2\sin B\cos A}{2\sin A\sin B}$ | M1 | |
| $= \frac{\cos A}{\sin A}$ | A1 | |
| $= \cot A$ | A1 cso | (3 marks) |
## Part (c)
| Answer/Working | Marks | Notes |
|---|---|---|
| Let $A = 75°$ and $B = 15°$ | B1 | |
| $\frac{\sin 90° - \sin 60°}{\cos 60° - \cos 90°} = \cot 75°$ | M1 | |
| $\cot 75° = \frac{1 - \frac{\sqrt{3}}{2}}{\frac{1}{2} - 0} = 2 - \sqrt{3}$ | M1 A1 | (4 marks) |
**Total: 11 marks**
---5. (a) Using the formulae
$$\begin{gathered}
\sin ( A \pm B ) = \sin A \cos B \pm \cos A \sin B \\
\cos ( A \pm B ) = \cos A \cos B \mp \sin A \sin B
\end{gathered}$$
show that
\begin{enumerate}[label=(\roman*)]
\item $\sin ( A + B ) - \sin ( A - B ) = 2 \cos A \sin B$,
\item $\cos ( A - B ) - \cos ( A + B ) = 2 \sin A \sin B$.\\
(b) Use the above results to show that
$$\frac { \sin ( A + B ) - \sin ( A - B ) } { \cos ( A - B ) - \cos ( A + B ) } = \cot A$$
Using the result of part (b) and the exact values of $\sin 60 ^ { \circ }$ and $\cos 60 ^ { \circ }$,\\
(c) find an exact value for $\cot 75 ^ { \circ }$ in its simplest form.\\
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5. continued & Leave blank \\
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\hfill \mbox{\textit{Edexcel C3 Q5 [11]}}