| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Show constant value or identity |
| Difficulty | Standard +0.3 Part (i) requires expanding two expressions and applying the Pythagorean identity sin²θ + cos²θ = 1, which is straightforward algebra. Part (ii) involves solving a linear trigonometric equation (2sinθ - 6cosθ = 3sinθ + cosθ) leading to tanθ = 7, which is routine. This is slightly easier than average as it's a standard textbook exercise testing basic identities and equation solving with no novel insight required. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(a^2 + b^2 = (s^2 + 9c^2 - 6sc) + (9s^2 + c^2 + 6sc)\) → \(10c^2 + 10s^2 = 10\) | B1 M1 A1 | Correct squaring. Use of \(s^2 + c^2 = 1\) to get constant. (can get 2/3 for missing \(6sc\)) |
| (ii) \(2s - 6c = 3s + c \rightarrow s = -7c\) → \(\tan\theta = -7\) → \(98.1°\) and \(278.1°\) | M1 A1 A1 A1 | Collecting and \(t = s+c\). For \(180° +\) first answer, providing no extra answers in the range |
| [4] |
$a = \sin\theta - 3\cos\theta$, $b = 3\sin\theta + \cos\theta$
(i) $a^2 + b^2 = (s^2 + 9c^2 - 6sc) + (9s^2 + c^2 + 6sc)$ → $10c^2 + 10s^2 = 10$ | B1 M1 A1 | Correct squaring. Use of $s^2 + c^2 = 1$ to get constant. (can get 2/3 for missing $6sc$)
(ii) $2s - 6c = 3s + c \rightarrow s = -7c$ → $\tan\theta = -7$ → $98.1°$ and $278.1°$ | M1 A1 A1 A1 | Collecting and $t = s+c$. For $180° +$ first answer, providing no extra answers in the range
[4] |
---5 It is given that $a = \sin \theta - 3 \cos \theta$ and $b = 3 \sin \theta + \cos \theta$, where $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.\\
(i) Show that $a ^ { 2 } + b ^ { 2 }$ has a constant value for all values of $\theta$.\\
(ii) Find the values of $\theta$ for which $2 a = b$.
\hfill \mbox{\textit{CAIE P1 2013 Q5 [7]}}