| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Convert equation to quadratic form |
| Difficulty | Moderate -0.3 This is a straightforward two-part question requiring standard manipulation of trig identities (tan θ = sin θ/cos θ, then sin²θ + cos²θ = 1) to reach a given quadratic form, followed by routine quadratic solving and finding angles. The 'show that' structure removes problem-solving challenge, making it slightly easier than average despite requiring multiple steps. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(4\cos^2\theta + 15\sin\theta = 0\) | M1 | Replace \(\tan\theta\) by \(\frac{\sin\theta}{\cos\theta}\) and multiply by \(\sin\theta\) or equivalent |
| \(4(1-s^2)+15s = 0 \rightarrow 4\sin^2\theta - 15\sin\theta - 4 = 0\) | M1A1 | Use \(c^2 = 1 - s^2\) and rearrange to AG (www) |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sin\theta = -\frac{1}{4}\) | B1 | Ignore other solution |
| \(\theta = 194.5\) or \(345.5\) | B1B1\(\sqrt{}\) | Ft from 1st solution, SC B1 both angles in rads (3.39 and 6.03) |
| Total: 3 |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4\cos^2\theta + 15\sin\theta = 0$ | M1 | Replace $\tan\theta$ by $\frac{\sin\theta}{\cos\theta}$ and multiply by $\sin\theta$ or equivalent |
| $4(1-s^2)+15s = 0 \rightarrow 4\sin^2\theta - 15\sin\theta - 4 = 0$ | M1A1 | Use $c^2 = 1 - s^2$ and rearrange to AG (www) |
| **Total: 3** | | |
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## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sin\theta = -\frac{1}{4}$ | B1 | Ignore other solution |
| $\theta = 194.5$ or $345.5$ | B1B1$\sqrt{}$ | Ft from 1st solution, SC B1 both angles in rads (3.39 and 6.03) |
| **Total: 3** | | |
---4 (i) Show that the equation $\frac { 4 \cos \theta } { \tan \theta } + 15 = 0$ can be expressed as
$$4 \sin ^ { 2 } \theta - 15 \sin \theta - 4 = 0$$
(ii) Hence solve the equation $\frac { 4 \cos \theta } { \tan \theta } + 15 = 0$ for $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.\\
\hfill \mbox{\textit{CAIE P1 Q4 [6]}}