| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Double angle with reciprocal functions |
| Difficulty | Challenging +1.2 Part (a) is a standard reciprocal trig identity proof requiring manipulation of cot and double angle formulas—routine for C3/C4 students. Part (b) requires recognizing the result from (a) to transform the equation, then solving a linear equation in the argument, which is moderately challenging but follows a clear path once the connection is made. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\cot x - \cot 2x \equiv \frac{\cos x}{\sin x}-\frac{\cos 2x}{\sin 2x}\) | B1 | |
| \(\equiv \frac{\sin 2x\cos x - \cos 2x\sin x}{\sin x\sin 2x}\) | M1 | |
| \(\equiv \frac{\sin(2x-x)}{\sin x\sin 2x}\) | M1 | |
| \(\equiv \frac{\sin x}{\sin x\sin 2x}\equiv\frac{1}{\sin 2x}\equiv \text{cosec}\, 2x\) | M1 A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2x=3\theta+\frac{\pi}{3} \Rightarrow x=1.5\theta+\frac{\pi}{6}\) | B1 | |
| \(\cot\left(1.5\theta+\frac{\pi}{6}\right)=\frac{1}{\sqrt{3}} \Rightarrow \tan\left(1.5\theta+\frac{\pi}{6}\right)=\sqrt{3}\) | M1 | |
| \(\left(1.5\theta+\frac{\pi}{6}\right)=\frac{\pi}{3},\ \frac{4\pi}{3}\) | M1 | |
| \(\theta=\frac{\pi}{9},\ \frac{7\pi}{9}\) | A1, A1 |
# Question 7:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\cot x - \cot 2x \equiv \frac{\cos x}{\sin x}-\frac{\cos 2x}{\sin 2x}$ | B1 | |
| $\equiv \frac{\sin 2x\cos x - \cos 2x\sin x}{\sin x\sin 2x}$ | M1 | |
| $\equiv \frac{\sin(2x-x)}{\sin x\sin 2x}$ | M1 | |
| $\equiv \frac{\sin x}{\sin x\sin 2x}\equiv\frac{1}{\sin 2x}\equiv \text{cosec}\, 2x$ | M1 A1* | |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2x=3\theta+\frac{\pi}{3} \Rightarrow x=1.5\theta+\frac{\pi}{6}$ | B1 | |
| $\cot\left(1.5\theta+\frac{\pi}{6}\right)=\frac{1}{\sqrt{3}} \Rightarrow \tan\left(1.5\theta+\frac{\pi}{6}\right)=\sqrt{3}$ | M1 | |
| $\left(1.5\theta+\frac{\pi}{6}\right)=\frac{\pi}{3},\ \frac{4\pi}{3}$ | M1 | |
| $\theta=\frac{\pi}{9},\ \frac{7\pi}{9}$ | A1, A1 | |
---\begin{enumerate}
\item (a) Show that
\end{enumerate}
$$\cot x - \cot 2 x \equiv \operatorname { cosec } 2 x , \quad x \neq \frac { n \pi } { 2 } , \quad n \in \mathbb { Z }$$
(b) Hence, or otherwise, solve for $0 \leqslant \theta \leqslant \pi$
$$\operatorname { cosec } \left( 3 \theta + \frac { \pi } { 3 } \right) + \cot \left( 3 \theta + \frac { \pi } { 3 } \right) = \frac { 1 } { \sqrt { 3 } }$$
You must show your working.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)\\
\hfill \mbox{\textit{Edexcel C34 Q7 [10]}}