| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Proofs |
| Type | Solve equation using proven identity |
| Difficulty | Moderate -0.3 This is a straightforward trigonometric identity proof requiring standard manipulations (converting tan to sin/cos, simplifying algebraically) followed by a routine equation solve using the proven identity. The proof involves 2-3 steps of algebraic manipulation with basic trig identities, and the equation solving is direct substitution leading to a simple quadratic in sin θ. Slightly easier than average due to its mechanical nature and standard techniques. |
| 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) \(\frac{\cos\theta}{\tan\theta(1 - \sin\theta)} = \frac{\cos^2\theta}{\sin\theta(1 - \sin\theta)}\) | M1 | Use of \(t = s ÷ c\) |
| \(= \frac{1 - \sin^2\theta}{\sin\theta(1 - \sin\theta)}\) | M1 | Replaces \(\cos^2\theta\) with \(1 - \sin^2\theta\) to form \(f(\sin\theta)\). |
| \(= \frac{1 + \sin\theta}{\sin\theta} = \frac{1}{\sin\theta} + 1\) | A1 [3] | AG. Ensure all ok. Must show difference of 2 squares. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\to \sin\theta = \frac{1}{3} \to \theta = 19.5°, 160.5°\) | M1 A1 A1∨ [3] | Linking up to obtain \(\sin\theta = k\). co \(\sqrt{180° - 1°}\) answer providing there are no other solutions in the range 0° to 360°. |
**(i)** $\frac{\cos\theta}{\tan\theta(1 - \sin\theta)} = \frac{\cos^2\theta}{\sin\theta(1 - \sin\theta)}$ | M1 | Use of $t = s ÷ c$
$= \frac{1 - \sin^2\theta}{\sin\theta(1 - \sin\theta)}$ | M1 | Replaces $\cos^2\theta$ with $1 - \sin^2\theta$ to form $f(\sin\theta)$.
$= \frac{1 + \sin\theta}{\sin\theta} = \frac{1}{\sin\theta} + 1$ | A1 [3] | AG. Ensure all ok. Must show difference of 2 squares.
**(ii)** $\frac{\cos\theta}{\tan\theta(1 - \sin\theta)} = 4 \to \frac{1}{\sin\theta} + 1 = 4$
$\to \sin\theta = \frac{1}{3} \to \theta = 19.5°, 160.5°$ | M1 A1 A1∨ [3] | Linking up to obtain $\sin\theta = k$. co $\sqrt{180° - 1°}$ answer providing there are no other solutions in the range 0° to 360°.\begin{enumerate}[label=(\roman*)]
\item Prove the identity $\frac{\cos \theta}{\tan \theta(1 - \sin \theta)} \equiv 1 + \frac{1}{\sin \theta}$. [3]
\item Hence solve the equation $\frac{\cos \theta}{\tan \theta(1 - \sin \theta)} = 4$, for $0° \leq \theta \leq 360°$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2011 Q5 [6]}}