| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Convert equation to quadratic form |
| Difficulty | Standard +0.3 This is a standard algebraic manipulation question requiring cross-multiplication, use of sin²θ + cos²θ = 1, and solving a quadratic. While it involves multiple steps, the techniques are routine for P1 level with no novel insight required. Slightly above average difficulty due to the algebraic complexity, but well within expected A-level scope. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals1.05p Proof involving trig: functions and identities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{(\cos\theta - 4)(5\cos\theta - 2) - 4\sin^2\theta}{\sin\theta(5\cos\theta - 2)} = 0\) | M1 | Accept numerator only |
| \(\frac{5\cos^2\theta - 22\cos\theta + 8 - 4(1-\cos^2\theta)}{\sin\theta(5\cos\theta-2)} = 0\) | M1 | Simplify numerator and use \(s^2 = 1 - c^2\). Accept numerator only |
| \(9\cos^2\theta - 22\cos\theta + 4 = 0\) AG | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt to solve for \(\cos\theta\) (formula, completing square expected) | M1 | Expect \(\cos\theta = 0.1978\). Allow 2.247 in addition |
| \(\theta = 78.6°, 281.4°\) (only, second solution in the range) | A1A1FT | Ft for \((360° - \text{1st solution})\) |
| Total: 3 |
## Question 5:
**Part (i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{(\cos\theta - 4)(5\cos\theta - 2) - 4\sin^2\theta}{\sin\theta(5\cos\theta - 2)} = 0$ | M1 | Accept numerator only |
| $\frac{5\cos^2\theta - 22\cos\theta + 8 - 4(1-\cos^2\theta)}{\sin\theta(5\cos\theta-2)} = 0$ | M1 | Simplify numerator and use $s^2 = 1 - c^2$. Accept numerator only |
| $9\cos^2\theta - 22\cos\theta + 4 = 0$ **AG** | A1 | |
| | **Total: 3** | |
**Part (ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt to solve for $\cos\theta$ (formula, completing square expected) | M1 | Expect $\cos\theta = 0.1978$. Allow 2.247 in addition |
| $\theta = 78.6°, 281.4°$ (only, second solution in the range) | A1A1FT | Ft for $(360° - \text{1st solution})$ |
| | **Total: 3** | |
---5 (i) Show that the equation
$$\frac { \cos \theta - 4 } { \sin \theta } - \frac { 4 \sin \theta } { 5 \cos \theta - 2 } = 0$$
may be expressed as $9 \cos ^ { 2 } \theta - 22 \cos \theta + 4 = 0$.\\
(ii) Hence solve the equation
$$\frac { \cos \theta - 4 } { \sin \theta } - \frac { 4 \sin \theta } { 5 \cos \theta - 2 } = 0$$
for $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.\\
\hfill \mbox{\textit{CAIE P1 2018 Q5 [6]}}