| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Prove identity then solve equation |
| Difficulty | Standard +0.3 This is a standard two-part trigonometric identity question requiring algebraic manipulation (finding common denominators, using difference of squares) followed by a routine equation solve. The techniques are straightforward for A-level: combine fractions, apply sin²θ + cos²θ = 1, then solve a quadratic in cos(2θ). Slightly above average due to the algebraic manipulation required, but well within typical P1 scope. |
| 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) \(\frac{\sin\theta(\sin\theta - \cos\theta) + \cos\theta(\sin\theta + \cos\theta)}{(\sin\theta + \cos\theta)(\sin\theta - \cos\theta)}\) | M1 | |
| \(\frac{\sin^2\theta - \sin\theta\cos\theta + \cos\theta\sin\theta = \cos^2\theta}{\sin^2\theta - \cos^2\theta}\) | A1 | |
| \(\frac{1}{\sin^2\theta - \cos^2\theta}\) | AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) \(s^2 - (1-s^2) = \frac{1}{3}\) or \(1 - c^2 - c^2 = \frac{1}{3}\) or \(3(s^2 - c^2) = c^2 + s^2\) | M1 | Applying \(c^2 + s^2 = 1\) |
| \(\sin\theta = (\pm)\sqrt{\frac{2}{3}}\) or \(\cos\theta = (\pm)\sqrt{\frac{1}{3}}\) or \(\tan\theta = (\pm)\sqrt{2}\) | A1 | Or \(s = (\pm) 0.816\), \(c = (\pm) 0.577\), \(t = (\pm) 1.414\) |
| \(\theta = 54.7°, 125.3°, 234.7°, 305.3°\) | A1A1 | any 2 solutions for 1st A1 >4 solutions in range max A1A0 |
(i) $\frac{\sin\theta(\sin\theta - \cos\theta) + \cos\theta(\sin\theta + \cos\theta)}{(\sin\theta + \cos\theta)(\sin\theta - \cos\theta)}$ | M1 |
$\frac{\sin^2\theta - \sin\theta\cos\theta + \cos\theta\sin\theta = \cos^2\theta}{\sin^2\theta - \cos^2\theta}$ | A1 |
$\frac{1}{\sin^2\theta - \cos^2\theta}$ | AG | A1 | www
[3]
(ii) $s^2 - (1-s^2) = \frac{1}{3}$ or $1 - c^2 - c^2 = \frac{1}{3}$ or $3(s^2 - c^2) = c^2 + s^2$ | M1 | Applying $c^2 + s^2 = 1$
$\sin\theta = (\pm)\sqrt{\frac{2}{3}}$ or $\cos\theta = (\pm)\sqrt{\frac{1}{3}}$ or $\tan\theta = (\pm)\sqrt{2}$ | A1 | Or $s = (\pm) 0.816$, $c = (\pm) 0.577$, $t = (\pm) 1.414$
$\theta = 54.7°, 125.3°, 234.7°, 305.3°$ | A1A1 | any 2 solutions for 1st A1 >4 solutions in range max A1A0
[4]5\\
(i) Show that $\frac { \sin \theta } { \sin \theta + \cos \theta } + \frac { \cos \theta } { \sin \theta - \cos \theta } \equiv \frac { 1 } { \sin ^ { 2 } \theta - \cos ^ { 2 } \theta }$.\\
(ii) Hence solve the equation $\frac { \sin \theta } { \sin \theta + \cos \theta } + \frac { \cos \theta } { \sin \theta - \cos \theta } = 3$, for $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.
\hfill \mbox{\textit{CAIE P1 2013 Q5 [7]}}