| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Expand compound angle then solve |
| Difficulty | Moderate -0.3 Part (a) requires straightforward application of inverse sine (taking sine of both sides, then solving a simple quadratic). Part (b) is a standard compound angle equation requiring knowledge of special angles and the general solution for sine. Both parts are routine A-level exercises with no novel problem-solving required, making this slightly easier than average. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(x^2 - 1 = \sin \frac{\pi}{3}\) | M1 | |
| \(x = \pm 1.366\) | A1A1♦ | ♦ for negative of 1st answer |
| [3] | ||
| (b) \(2\theta + \frac{\pi}{3} = \frac{5\pi}{6}\) (or \(\frac{13\pi}{6}\) or \(\frac{\pi}{6}\)) | B1 | 1 correct angle on RHS is sufficient |
| \(2\theta = \frac{\pi}{2}\) or (\(\frac{11\pi}{6}\)) | M1 | Isolating \(2\theta\) |
| \(\theta = \frac{\pi}{4}, \frac{11\pi}{12}\) | A1A1 | SC decimals 0.785 & 2.88 scores M1B1 |
| [4] |
**(a)** $x^2 - 1 = \sin \frac{\pi}{3}$ | M1 |
$x = \pm 1.366$ | A1A1♦ | ♦ for negative of 1st answer
| | [3]
**(b)** $2\theta + \frac{\pi}{3} = \frac{5\pi}{6}$ (or $\frac{13\pi}{6}$ or $\frac{\pi}{6}$) | B1 | 1 correct angle on RHS is sufficient
$2\theta = \frac{\pi}{2}$ or ($\frac{11\pi}{6}$) | M1 | Isolating $2\theta$
$\theta = \frac{\pi}{4}, \frac{11\pi}{12}$ | A1A1 | SC decimals 0.785 & 2.88 scores M1B1
| | [4]7
\begin{enumerate}[label=(\alph*)]
\item Find the possible values of $x$ for which $\sin ^ { - 1 } \left( x ^ { 2 } - 1 \right) = \frac { 1 } { 3 } \pi$, giving your answers correct to 3 decimal places.
\item Solve the equation $\sin \left( 2 \theta + \frac { 1 } { 3 } \pi \right) = \frac { 1 } { 2 }$ for $0 \leqslant \theta \leqslant \pi$, giving $\theta$ in terms of $\pi$ in your answers.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2013 Q7 [7]}}