| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove identity then solve equation |
| Difficulty | Standard +0.8 This question requires proving a non-trivial reciprocal trig identity using double angle formulas and cotangent/tangent relationships, then applying it to solve an equation involving cosec². The proof demands algebraic manipulation of multiple trig identities, and the 'hence' part requires recognizing how to use the proven identity with the given equation. While systematic, it's above average difficulty due to the reciprocal functions and multi-step algebraic reasoning required. |
| 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 | Mark | Guidance |
| \(2\cot 2x + \tan x \equiv \dfrac{2}{\tan 2x} + \tan x\) | B1 | States or uses the identity \(2\cot 2x = \dfrac{2}{\tan 2x}\) or \(2\cot 2x = \dfrac{2\cos 2x}{\sin 2x}\). Note \(2\cot 2x = \dfrac{1}{2\tan 2x}\) is B0 |
| \(\equiv \dfrac{(1-\tan^2 x)}{\tan x} + \dfrac{\tan^2 x}{\tan x}\) | M1 | Uses the correct double angle identity \(\tan 2x = \dfrac{2\tan x}{1-\tan^2 x}\). Alternatively uses \(\sin 2x = 2\sin x\cos x\), \(\cos 2x = \cos^2 x - \sin^2 x\) and \(\tan x = \dfrac{\sin x}{\cos x}\) |
| \(\equiv \dfrac{1}{\tan x}\) | M1 | Writes two terms with a single common denominator and simplifies to a form \(\dfrac{ab}{cd}\) in either \(\sin x\) and \(\cos x\) or just \(\tan x\) |
| \(\equiv \cot x\) | A1* | cso. For proceeding to the correct answer — all aspects must be correct including consistent use of variables. If approaching from both sides there must be a conclusion for this mark. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(6\cot 2x + 3\tan x = \cosec^2 x - 2 \Rightarrow 3\cot x = \cosec^2 x - 2\) | M1 | For using part (a) and writing \(6\cot 2x + 3\tan x\) as \(k\cot x\), \(k\neq 0\), WITH an attempt at using \(\cosec^2 x = \pm 1 \pm \cot^2 x\) to produce a quadratic in just \(\cot x / \tan x\) |
| \(\Rightarrow 3\cot x = 1 + \cot^2 x - 2\) | ||
| \(\Rightarrow 0 = \cot^2 x - 3\cot x - 1\) | A1 | \(\cot^2 x - 3\cot x - 1 = 0\). The \(=0\) may be implied by subsequent working. Alternatively accept \(\tan^2 x + 3\tan x - 1 = 0\) |
| \(\Rightarrow \cot x = \dfrac{3 \pm \sqrt{13}}{2}\) | M1 | Solves a 3TQ\(=0\) in \(\cot x\) (or \(\tan\)) using the formula or any suitable method to find at least one solution. FYI: \(\cot x \approx 3.30,\ -0.30\) |
| \(\Rightarrow \tan x = \dfrac{2}{3 \pm \sqrt{13}} \Rightarrow x = \ldots\) | M1 | For \(\tan x = \dfrac{1}{\cot x}\) and using arctan producing at least one answer for \(x\) in degrees or radians |
| \(\Rightarrow x = 0.294,\ -2.848,\ -1.277,\ 1.865\) | A1 | Two of \(x = 0.294,\ -2.848,\ -1.277,\ 1.865\) (awrt 3dp) in radians or degrees. In degrees: \(x = 16.8°,\ 106.8°,\ -73.2°,\ -163.2°\) (awrt 2dp) |
| A1 | All four of \(x = 0.294,\ -2.848,\ -1.277,\ 1.865\) (awrt 3dp) with no extra solutions in range \(-\pi \leq x \leq \pi\) |
## Question 8:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\cot 2x + \tan x \equiv \dfrac{2}{\tan 2x} + \tan x$ | B1 | States or uses the identity $2\cot 2x = \dfrac{2}{\tan 2x}$ or $2\cot 2x = \dfrac{2\cos 2x}{\sin 2x}$. Note $2\cot 2x = \dfrac{1}{2\tan 2x}$ is B0 |
| $\equiv \dfrac{(1-\tan^2 x)}{\tan x} + \dfrac{\tan^2 x}{\tan x}$ | M1 | Uses the correct double angle identity $\tan 2x = \dfrac{2\tan x}{1-\tan^2 x}$. Alternatively uses $\sin 2x = 2\sin x\cos x$, $\cos 2x = \cos^2 x - \sin^2 x$ and $\tan x = \dfrac{\sin x}{\cos x}$ |
| $\equiv \dfrac{1}{\tan x}$ | M1 | Writes two terms with a single common denominator and simplifies to a form $\dfrac{ab}{cd}$ in either $\sin x$ and $\cos x$ or just $\tan x$ |
| $\equiv \cot x$ | A1* | cso. For proceeding to the correct answer — all aspects must be correct including consistent use of variables. If approaching from both sides there must be a conclusion for this mark. |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $6\cot 2x + 3\tan x = \cosec^2 x - 2 \Rightarrow 3\cot x = \cosec^2 x - 2$ | M1 | For using part (a) and writing $6\cot 2x + 3\tan x$ as $k\cot x$, $k\neq 0$, WITH an attempt at using $\cosec^2 x = \pm 1 \pm \cot^2 x$ to produce a quadratic in just $\cot x / \tan x$ |
| $\Rightarrow 3\cot x = 1 + \cot^2 x - 2$ | | |
| $\Rightarrow 0 = \cot^2 x - 3\cot x - 1$ | A1 | $\cot^2 x - 3\cot x - 1 = 0$. The $=0$ may be implied by subsequent working. Alternatively accept $\tan^2 x + 3\tan x - 1 = 0$ |
| $\Rightarrow \cot x = \dfrac{3 \pm \sqrt{13}}{2}$ | M1 | Solves a 3TQ$=0$ in $\cot x$ (or $\tan$) using the formula or any suitable method to find at least one solution. FYI: $\cot x \approx 3.30,\ -0.30$ |
| $\Rightarrow \tan x = \dfrac{2}{3 \pm \sqrt{13}} \Rightarrow x = \ldots$ | M1 | For $\tan x = \dfrac{1}{\cot x}$ and using arctan producing at least one answer for $x$ in degrees or radians |
| $\Rightarrow x = 0.294,\ -2.848,\ -1.277,\ 1.865$ | A1 | Two of $x = 0.294,\ -2.848,\ -1.277,\ 1.865$ (awrt 3dp) in radians or degrees. In degrees: $x = 16.8°,\ 106.8°,\ -73.2°,\ -163.2°$ (awrt 2dp) |
| | A1 | All four of $x = 0.294,\ -2.848,\ -1.277,\ 1.865$ (awrt 3dp) with no extra solutions in range $-\pi \leq x \leq \pi$ |
---\begin{enumerate}
\item (a) Prove that
\end{enumerate}
$$2 \cot 2 x + \tan x \equiv \cot x \quad x \neq \frac { n \pi } { 2 } , n \in \mathbb { Z }$$
(b) Hence, or otherwise, solve, for $- \pi \leqslant x < \pi$,
$$6 \cot 2 x + 3 \tan x = \operatorname { cosec } ^ { 2 } x - 2$$
Give your answers to 3 decimal places.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)
\includegraphics[max width=\textwidth, alt={}, center]{d3ba2776-eedb-48f0-834f-41aa454afba3-14_2258_47_315_37}\\
\hfill \mbox{\textit{Edexcel C3 2016 Q8 [10]}}