| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2016 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Solve equation with reciprocal functions |
| Difficulty | Standard +0.3 This is a straightforward application of the Pythagorean identity cot²x + 1 = cosec²x to transform the equation, followed by solving a quadratic in cosec x and finding angles. It requires standard technique with reciprocal trig functions but involves no novel insight—slightly easier than average due to the guided 'show that' part and routine quadratic solving. |
| Spec | 1.05k 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 |
| \(\text{cosec}^2x - \text{cosec}\,x - 12 = 0\) or \(k=-12\) | B1 | No working required; if \(k=12\) written with equation, allow isw |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((\text{cosec}\,x-4)(\text{cosec}\,x+3)=0 \Rightarrow \text{cosec}\,x = \ldots\) | M1 | Solves quadratic in cosec \(x\) by any method; correct answers (cosec \(x=4\) and \(-3\)) imply M1 |
| \(\sin x = \frac{1}{4}\) or \(-\frac{1}{3}\) | dM1 | Uses \(\sin x = \frac{1}{\text{cosec}\,x}\); dependent on first M1 |
| One answer in range \([0°,360°]\) from arcsin | dM1 | For using arcsin to produce one answer in range \(0\) to \(360°\) |
| \(x = 14.5°\) or \(165.5°\) or \(340.5°\) or \(199.5°\) | dM1, A1, A1 | dM1+A1: two correct answers in range; A1: all four correct; extra solutions withhold last A1; ignore solutions outside \(0\leq x \leq 360°\) |
# Question 2:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{cosec}^2x - \text{cosec}\,x - 12 = 0$ or $k=-12$ | B1 | No working required; if $k=12$ written with equation, allow isw |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(\text{cosec}\,x-4)(\text{cosec}\,x+3)=0 \Rightarrow \text{cosec}\,x = \ldots$ | M1 | Solves quadratic in cosec $x$ by any method; correct answers (cosec $x=4$ and $-3$) imply M1 |
| $\sin x = \frac{1}{4}$ or $-\frac{1}{3}$ | dM1 | Uses $\sin x = \frac{1}{\text{cosec}\,x}$; dependent on first M1 |
| One answer in range $[0°,360°]$ from arcsin | dM1 | For using arcsin to produce one answer in range $0$ to $360°$ |
| $x = 14.5°$ or $165.5°$ or $340.5°$ or $199.5°$ | dM1, A1, A1 | dM1+A1: two correct answers in range; A1: all four correct; extra solutions withhold last A1; ignore solutions outside $0\leq x \leq 360°$ |
---\begin{enumerate}
\item (a) Show that
\end{enumerate}
$$\cot ^ { 2 } x - \operatorname { cosec } x - 11 = 0$$
may be expressed in the form $\operatorname { cosec } ^ { 2 } x - \operatorname { cosec } x + k = 0$, where $k$ is a constant.\\
(b) Hence solve for $0 \leqslant x < 360 ^ { \circ }$
$$\cot ^ { 2 } x - \operatorname { cosec } x - 11 = 0$$
Give each solution in degrees to one decimal place.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)\\
\hfill \mbox{\textit{Edexcel C34 2016 Q2 [6]}}