| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2014 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Double angle with reciprocal functions |
| Difficulty | Standard +0.3 This is a standard C3/C4 reciprocal trig identity proof followed by routine equation solving. Part (a) requires converting reciprocal functions to sin/cos and using the double angle formula—a common textbook exercise. Parts (b)(i) and (b)(ii) are direct applications requiring substitution and basic algebraic manipulation. While it tests multiple techniques, it follows predictable patterns without requiring novel insight, making it slightly easier than average. |
| Spec | 1.05j 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 |
| \(2\cosec 2A = \frac{2}{\sin 2A}\) | B1 | Writes \(2\cosec 2A\) as \(\frac{2}{\sin 2A}\) |
| \(= \frac{2}{2\sin A\cos A} - \frac{\cos A}{\sin A}\) | M1 | Uses double angle formula for \(\sin 2A\) and \(\cot A = \frac{\cos A}{\sin A}\); accept \(\frac{2}{2\sin A\cos A}\) or \(\frac{2}{\sin A\cos A + \cos A\sin A}\) |
| \(= \frac{2 - 2\cos^2 A}{2\sin A\cos A}\) | M1 | Single fraction in terms of \(\sin A\) and \(\cos A\) only; denominator correct, at least one numerator modified |
| \(\frac{2(1-\cos^2 A)}{2\sin A\cos A} = \frac{2\sin^2 A}{2\sin A\cos A} = \frac{\sin A}{\cos A} = \tan A\) | A1* | Completely correct proof; \(1-\cos^2 A\) must be replaced with \(\sin^2 A\); \(\frac{\sin A}{\cos A}\) must be clearly seen before replacing with \(\tan A\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2\cosec 4\theta - \cot 2\theta = \sqrt{3} \Rightarrow \tan 2\theta = \sqrt{3}\) | M1 | Uses part (a) to write equation in form \(\tan 2\theta = \sqrt{3}\); may be implied by \(A = 2\theta\) |
| \(\theta = \frac{\arctan\sqrt{3}}{2} = \frac{\pi}{6}\), accept awrt \(0.524\) | A1 | \(\theta = \frac{\pi}{6}\); not \(30°\); ignore extra solutions outside range; withhold if extra solutions given inside range |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\tan\theta + \cot\theta = 5 \Rightarrow \cosec 2\theta = \frac{5}{2}\) | M1 | Uses part (a) to write equation in form \(\cosec 2\theta = C\) |
| \(\theta = \frac{1}{2}\arcsin\!\left(\frac{2}{5}\right) =\) awrt \(0.206, 1.37\) | dM1, A1, A1 | dM1: correct order of operations \(\cosec 2\theta = C \Rightarrow \sin 2\theta = \frac{1}{C} \Rightarrow 2\theta = \arcsin\!\left(\frac{1}{C}\right) \Rightarrow \theta = \ldots\); A1 one correct solution awrt 0.206 or 1.37; A1 both solutions correct, no extras inside range |
## Question 8:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2\cosec 2A = \frac{2}{\sin 2A}$ | B1 | Writes $2\cosec 2A$ as $\frac{2}{\sin 2A}$ |
| $= \frac{2}{2\sin A\cos A} - \frac{\cos A}{\sin A}$ | M1 | Uses double angle formula for $\sin 2A$ and $\cot A = \frac{\cos A}{\sin A}$; accept $\frac{2}{2\sin A\cos A}$ or $\frac{2}{\sin A\cos A + \cos A\sin A}$ |
| $= \frac{2 - 2\cos^2 A}{2\sin A\cos A}$ | M1 | Single fraction in terms of $\sin A$ and $\cos A$ only; denominator correct, at least one numerator modified |
| $\frac{2(1-\cos^2 A)}{2\sin A\cos A} = \frac{2\sin^2 A}{2\sin A\cos A} = \frac{\sin A}{\cos A} = \tan A$ | A1* | Completely correct proof; $1-\cos^2 A$ must be replaced with $\sin^2 A$; $\frac{\sin A}{\cos A}$ must be clearly seen before replacing with $\tan A$ |
### Part (b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2\cosec 4\theta - \cot 2\theta = \sqrt{3} \Rightarrow \tan 2\theta = \sqrt{3}$ | M1 | Uses part (a) to write equation in form $\tan 2\theta = \sqrt{3}$; may be implied by $A = 2\theta$ |
| $\theta = \frac{\arctan\sqrt{3}}{2} = \frac{\pi}{6}$, accept awrt $0.524$ | A1 | $\theta = \frac{\pi}{6}$; not $30°$; ignore extra solutions outside range; withhold if extra solutions given inside range |
### Part (b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan\theta + \cot\theta = 5 \Rightarrow \cosec 2\theta = \frac{5}{2}$ | M1 | Uses part (a) to write equation in form $\cosec 2\theta = C$ |
| $\theta = \frac{1}{2}\arcsin\!\left(\frac{2}{5}\right) =$ awrt $0.206, 1.37$ | dM1, A1, A1 | dM1: correct order of operations $\cosec 2\theta = C \Rightarrow \sin 2\theta = \frac{1}{C} \Rightarrow 2\theta = \arcsin\!\left(\frac{1}{C}\right) \Rightarrow \theta = \ldots$; A1 one correct solution awrt 0.206 or 1.37; A1 both solutions correct, no extras inside range |
---8. (a) Prove that
$$\text { 2cosec } 2 A - \cot A \equiv \tan A , \quad A \neq \frac { n \pi } { 2 } , n \in \mathbb { Z }$$
(b) Hence solve, for $0 \leqslant \theta \leqslant \frac { \pi } { 2 }$
\begin{enumerate}[label=(\roman*)]
\item $2 \operatorname { cosec } 4 \theta - \cot 2 \theta = \sqrt { } 3$
\item $\tan \theta + \cot \theta = 5$
Give your answers to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2014 Q8 [10]}}