| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Double angle with reciprocal functions |
| Difficulty | Standard +0.8 This is a multi-part question requiring manipulation of reciprocal trig identities with double angles, followed by a specific numerical verification and solving a non-standard equation with half-angles. Part (i) is routine identity proof, part (ii) is straightforward application, but part (iii) requires substitution, algebraic manipulation of reciprocal functions with half-angles, and careful consideration of the extended domain, making it moderately challenging overall. |
| 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 |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(2\cosec 2\theta \cot\theta = \frac{2}{\sin 2\theta} \cdot \frac{\cos\theta}{\sin\theta}\) | M1 | Express in terms of sin/cos |
| \(= \frac{2\cos\theta}{2\sin\theta\cos\theta \cdot \sin\theta}\) | M1 | Use \(\sin 2\theta = 2\sin\theta\cos\theta\) |
| \(= \frac{1}{\sin^2\theta} = \cosec^2\theta\) | A1 | Completion |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Use result with \(\theta = 15°\): \(2\cosec 30° \cot 15° = \cosec^2 15°\) | M1 | Correct substitution |
| \(2 \times 2 \times \cot 15° = \cosec^2 15°\), so \(\cosec^2 15° \tan 15° = 4\) | A1 | Completion shown correctly |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Apply part (i) with \(\theta = \frac{\phi}{2}\): \(2\cosec\phi\cot\frac{\phi}{2} = \cosec^2\frac{\phi}{2}\) | M1 | |
| Equation becomes \(\cosec^2\frac{\phi}{2} + \cosec\frac{\phi}{2} = 12\) | M1 | |
| \(\cosec^2\frac{\phi}{2} + \cosec\frac{\phi}{2} - 12 = 0\) | M1 | Form quadratic |
| \((\cosec\frac{\phi}{2} + 4)(\cosec\frac{\phi}{2} - 3) = 0\) | M1 | Solve quadratic |
| \(\sin\frac{\phi}{2} = -\frac{1}{4}\) or \(\sin\frac{\phi}{2} = \frac{1}{3}\) | ||
| \(\phi = \pm 338.6°, \pm 179.4°\) (relevant values in range) | A1 | All correct solutions |
## Question 7(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $2\cosec 2\theta \cot\theta = \frac{2}{\sin 2\theta} \cdot \frac{\cos\theta}{\sin\theta}$ | M1 | Express in terms of sin/cos |
| $= \frac{2\cos\theta}{2\sin\theta\cos\theta \cdot \sin\theta}$ | M1 | Use $\sin 2\theta = 2\sin\theta\cos\theta$ |
| $= \frac{1}{\sin^2\theta} = \cosec^2\theta$ | A1 | Completion |
## Question 7(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Use result with $\theta = 15°$: $2\cosec 30° \cot 15° = \cosec^2 15°$ | M1 | Correct substitution |
| $2 \times 2 \times \cot 15° = \cosec^2 15°$, so $\cosec^2 15° \tan 15° = 4$ | A1 | Completion shown correctly |
## Question 7(iii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Apply part (i) with $\theta = \frac{\phi}{2}$: $2\cosec\phi\cot\frac{\phi}{2} = \cosec^2\frac{\phi}{2}$ | M1 | |
| Equation becomes $\cosec^2\frac{\phi}{2} + \cosec\frac{\phi}{2} = 12$ | M1 | |
| $\cosec^2\frac{\phi}{2} + \cosec\frac{\phi}{2} - 12 = 0$ | M1 | Form quadratic |
| $(\cosec\frac{\phi}{2} + 4)(\cosec\frac{\phi}{2} - 3) = 0$ | M1 | Solve quadratic |
| $\sin\frac{\phi}{2} = -\frac{1}{4}$ or $\sin\frac{\phi}{2} = \frac{1}{3}$ | | |
| $\phi = \pm 338.6°, \pm 179.4°$ (relevant values in range) | A1 | All correct solutions |
This image shows an **Additional Page** (page 12) from a Cambridge A-Level Mathematics exam paper (9709/21/M/J/19). It contains only blank lined space for students to write additional answers — there is **no mark scheme content** on this page.
This is a **question paper**, not a mark scheme. To extract mark scheme content, you would need to provide the actual mark scheme document for this paper (9709/21/M/J/19).7 (i) Show that $2 \operatorname { cosec } 2 \theta \cot \theta \equiv \operatorname { cosec } ^ { 2 } \theta$.\\
(ii) Hence show that $\operatorname { cosec } ^ { 2 } 15 ^ { \circ } \tan 15 ^ { \circ } = 4$.\\
(iii) Solve the equation $2 \operatorname { cosec } \phi \cot \frac { 1 } { 2 } \phi + \operatorname { cosec } \frac { 1 } { 2 } \phi = 12$ for $- 360 ^ { \circ } < \phi < 360 ^ { \circ }$. Show all necessary working.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE P2 2019 Q7 [10]}}