| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Convert equation to quadratic form |
| Difficulty | Moderate -0.3 This is a standard A-level technique of converting a trigonometric equation to quadratic form using the Pythagorean identity sin²θ + cos²θ = 1. Part (i) is straightforward algebraic manipulation (substituting sin²θ = 1 - cos²θ), and part (ii) requires solving a simple quadratic then finding angles from cos²θ values. The question is slightly easier than average as it's a well-practiced exam technique with clear guidance ('show that' and 'hence'), though it does require multiple steps and careful handling of the squared cosine values. |
| 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 |
| \(3\cos^4\theta + 4(1 - \cos^2\theta) - 3\ (=0)\) | M1 | Use \(s^2 = 1 - c^2\) |
| \(3x^2 + 4(1-x) - 3(=0) \rightarrow 3x^2 - 4x + 1\ (=0)\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt to solve for \(x\) | M1 | Expect \(x = 1,\ 1/3\) |
| \(\cos\theta = (\pm)1,\ (\pm)0.5774\) | A1 | Accept \((\pm)\left(\frac{1}{\sqrt{3}}\right)\) SOI |
| \((\theta =)\ 0°,\ 180°,\ 54.7°,\ 125.3°\) | A3,2,1,0 | A2,1,0 if more than 4 solutions in range |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3\cos^4\theta + 4(1 - \cos^2\theta) - 3\ (=0)$ | M1 | Use $s^2 = 1 - c^2$ |
| $3x^2 + 4(1-x) - 3(=0) \rightarrow 3x^2 - 4x + 1\ (=0)$ | A1 | AG |
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt to solve for $x$ | M1 | Expect $x = 1,\ 1/3$ |
| $\cos\theta = (\pm)1,\ (\pm)0.5774$ | A1 | Accept $(\pm)\left(\frac{1}{\sqrt{3}}\right)$ SOI |
| $(\theta =)\ 0°,\ 180°,\ 54.7°,\ 125.3°$ | A3,2,1,0 | A2,1,0 if more than 4 solutions in range |
---7 (i) Show that the equation $3 \cos ^ { 4 } \theta + 4 \sin ^ { 2 } \theta - 3 = 0$ can be expressed as $3 x ^ { 2 } - 4 x + 1 = 0$, where $x = \cos ^ { 2 } \theta$.\\
(ii) Hence solve the equation $3 \cos ^ { 4 } \theta + 4 \sin ^ { 2 } \theta - 3 = 0$ for $0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }$.\\
\hfill \mbox{\textit{CAIE P1 2019 Q7 [7]}}