| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Quadratic trigonometric equations |
| Type | Quartic in sin or cos substitution |
| Difficulty | Moderate -0.3 Part (i) requires routine algebraic manipulation using tan²θ = sin²θ/cos²θ and the Pythagorean identity, which is standard bookwork. Part (ii) involves solving a quadratic in sin²θ and finding angles in a given range—straightforward application of techniques. The question is slightly easier than average due to its structured guidance and use of well-practiced identities. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{2\sin^2 \theta - 6\sin^2 \theta}{1 - \sin^2 \theta} = 1\) | M1 | Equation as function of sin θ |
| \(2\sin^2 \theta + \sin \theta - 1 = 0\) | AG | A1 [2] |
| (ii) \((2\sin^2 \theta - 1)(\sin \theta + 1) = 0\) | M1 | Or use formula on quadratic in sin² θ |
| \(\sin \theta = \frac{(\pm)1}{\sqrt{2}}\) | A1 | |
| \(\theta = 45°, 135°\) | A1 | |
| \(\theta = 225°, 315°\) | A1 [4] | Provided no excess solutions in range |
**(i)** $\frac{2\sin^2 \theta - 6\sin^2 \theta}{1 - \sin^2 \theta} = 1$ | M1 | Equation as function of sin θ
$2\sin^2 \theta + \sin \theta - 1 = 0$ | AG | A1 [2] |
**(ii)** $(2\sin^2 \theta - 1)(\sin \theta + 1) = 0$ | M1 | Or use formula on quadratic in sin² θ
$\sin \theta = \frac{(\pm)1}{\sqrt{2}}$ | A1 |
$\theta = 45°, 135°$ | A1 |
$\theta = 225°, 315°$ | A1 [4] | Provided no excess solutions in range
---5 (i) Show that the equation $2 \tan ^ { 2 } \theta \sin ^ { 2 } \theta = 1$ can be written in the form
$$2 \sin ^ { 4 } \theta + \sin ^ { 2 } \theta - 1 = 0 .$$
(ii) Hence solve the equation $2 \tan ^ { 2 } \theta \sin ^ { 2 } \theta = 1$ for $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.
\hfill \mbox{\textit{CAIE P1 2011 Q5 [6]}}