| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Convert equation to quadratic form |
| Difficulty | Moderate -0.3 This is a slightly below-average A-level question. Part (i) requires the standard identity tan²θ = sin²θ/cos²θ and sin²θ = 1 - cos²θ to convert to quadratic form—straightforward algebraic manipulation. Part (ii) involves solving the resulting quadratic and applying inverse trig, but the restricted domain and exact answers in terms of π keep it routine. No novel insight required, just systematic application of standard identities. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\rightarrow 2\cos^2\theta = \frac{\sin^2\theta}{\cos^2\theta}\) | M1 | Use of \(t^2 = s^2 \div c^2\) or alternative. Correct eqn |
| \(\rightarrow\) Uses \(c^2 + s^2 = 1 \rightarrow 2c^4 = 1 - c^2\) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((2c^2-1)(c^2+1) = 0 \rightarrow c = \pm\frac{1}{\sqrt{2}}\) | M1 | Method of solving for 3-term quadratic |
| \(\rightarrow \theta = \frac{1}{4}\pi\) or \(\frac{3}{4}\pi\) | A1 A1\(\sqrt{}\) [3] | (in terms of \(\pi\)). \(\sqrt{}\) for \(\pi - 1\)st ans. Cannot gain A1\(\sqrt{}\) if other answers given in the range |
## Question 3:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\rightarrow 2\cos^2\theta = \frac{\sin^2\theta}{\cos^2\theta}$ | M1 | Use of $t^2 = s^2 \div c^2$ or alternative. Correct eqn |
| $\rightarrow$ Uses $c^2 + s^2 = 1 \rightarrow 2c^4 = 1 - c^2$ | A1 [2] | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(2c^2-1)(c^2+1) = 0 \rightarrow c = \pm\frac{1}{\sqrt{2}}$ | M1 | Method of solving for 3-term quadratic |
| $\rightarrow \theta = \frac{1}{4}\pi$ or $\frac{3}{4}\pi$ | A1 A1$\sqrt{}$ [3] | (in terms of $\pi$). $\sqrt{}$ for $\pi - 1$st ans. Cannot gain A1$\sqrt{}$ if other answers given in the range |
---3 (i) Express the equation $2 \cos ^ { 2 } \theta = \tan ^ { 2 } \theta$ as a quadratic equation in $\cos ^ { 2 } \theta$.\\
(ii) Solve the equation $2 \cos ^ { 2 } \theta = \tan ^ { 2 } \theta$ for $0 \leqslant \theta \leqslant \pi$, giving solutions in terms of $\pi$.
\hfill \mbox{\textit{CAIE P1 2013 Q3 [5]}}