| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2004 |
| Session | June |
| Marks | 5 |
| 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 trigonometric equation requiring division by cos²θ to convert to tan θ, then solving a straightforward quadratic. The technique is commonly taught and practiced in P1, with no unusual insight required. Slightly above average difficulty due to the algebraic manipulation and range consideration, but well within typical A-level expectations. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks |
|---|---|
| M1 A1 [2] | Knowing to divide by \(\cos^2 \theta\); Correct quadratic (not \(nec = 0\)) |
| Answer | Marks |
|---|---|
| M1 [1] | Correct solution of quadratic = 0 |
| Answer | Marks |
|---|---|
| A1 A1 [3] | Correct only for each one. |
**(i)** $\sin^2 \theta + 3\sin \theta \cos \theta = 4\cos^2 \theta$ divides by $\cos^2 \theta$ → $\tan^2 \theta + 3\tan \theta = 4$
| M1 A1 [2] | Knowing to divide by $\cos^2 \theta$; Correct quadratic (not $nec = 0$) |
**(ii)** Solution $\tan \theta = 1$ or $\tan \theta = -4$
| M1 [1] | Correct solution of quadratic = 0 |
$\to \theta = 45°$ or $104.0°$
| A1 A1 [3] | Correct only for each one. |
---3 (i) Show that the equation $\sin ^ { 2 } \theta + 3 \sin \theta \cos \theta = 4 \cos ^ { 2 } \theta$ can be written as a quadratic equation in $\tan \theta$.\\
(ii) Hence, or otherwise, solve the equation in part (i) for $0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }$.
\hfill \mbox{\textit{CAIE P1 2004 Q3 [5]}}