| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2011 |
| 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 two-part question requiring algebraic manipulation of trig identities followed by a routine equation solve. Part (i) involves standard simplification using sin/cos/tan relationships and algebraic manipulation (difference of squares pattern). Part (ii) is straightforward once the identity is established: equate to 2/5, solve for cos θ, then find angles. While it requires careful algebra, it follows predictable patterns without requiring novel insight, making it slightly easier than average. |
| 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) \(\left(\frac{1}{\sin\theta} - \frac{1}{\tan\theta}\right)^2 = \frac{1-\cos\theta}{1+\cos\theta}\) | ||
| \(\left(\frac{1}{\sin\theta} - \frac{\cos\theta}{\sin\theta}\right)^2 = \frac{(1-\cos\theta)^2}{\sin^2\theta}\) | M1 | Use of \(\tan = \sin/\cos\) |
| \(= \frac{(1-\cos\theta)(1-\cos\theta)}{1-\cos^2\theta} = \frac{1-\cos\theta}{1+\cos\theta}\) | M1 A1 | Use of \(\sin^2 + \cos^2 = 1\). All correct. |
| [3] | (NB ag. – ensure cancelling has been done) | |
| (ii) \(\left(\frac{1}{\sin\theta} - \frac{1}{\tan\theta}\right)^2 = \frac{2}{5}\) | ||
| \(\frac{1-\cos\theta}{1+\cos\theta} = \frac{2}{5}\) | M1 | Uses part (i) to obtain an eqn in \(\cos\theta\) |
| \(\cos\theta = \frac{3}{7}\) | A1 | co |
| \(\theta = 64.6°\) or \(295.4°\) | A1 A1 √ | co. √ for 360 – "1st answer". |
| [4] |
(i) $\left(\frac{1}{\sin\theta} - \frac{1}{\tan\theta}\right)^2 = \frac{1-\cos\theta}{1+\cos\theta}$ | |
$\left(\frac{1}{\sin\theta} - \frac{\cos\theta}{\sin\theta}\right)^2 = \frac{(1-\cos\theta)^2}{\sin^2\theta}$ | M1 | Use of $\tan = \sin/\cos$
$= \frac{(1-\cos\theta)(1-\cos\theta)}{1-\cos^2\theta} = \frac{1-\cos\theta}{1+\cos\theta}$ | M1 A1 | Use of $\sin^2 + \cos^2 = 1$. All correct.
| [3] | (NB ag. – ensure cancelling has been done)
(ii) $\left(\frac{1}{\sin\theta} - \frac{1}{\tan\theta}\right)^2 = \frac{2}{5}$ | |
$\frac{1-\cos\theta}{1+\cos\theta} = \frac{2}{5}$ | M1 | Uses part (i) to obtain an eqn in $\cos\theta$
$\cos\theta = \frac{3}{7}$ | A1 | co
$\theta = 64.6°$ or $295.4°$ | A1 A1 √ | co. √ for 360 – "1st answer".
| [4] |8 (i) Prove the identity $\left( \frac { 1 } { \sin \theta } - \frac { 1 } { \tan \theta } \right) ^ { 2 } \equiv \frac { 1 - \cos \theta } { 1 + \cos \theta }$.\\
(ii) Hence solve the equation $\left( \frac { 1 } { \sin \theta } - \frac { 1 } { \tan \theta } \right) ^ { 2 } = \frac { 2 } { 5 }$, for $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.
\hfill \mbox{\textit{CAIE P1 2011 Q8 [7]}}