| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2016 |
| Session | March |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Inverse trigonometric function equations |
| Difficulty | Moderate -0.3 Part (a) requires direct application of inverse sine definition with exact value arithmetic. Part (b) involves factorising a trigonometric expression (grouping method) then solving basic equations, but is straightforward once factorised. Both parts are routine A-level techniques with no novel insight required, making this slightly easier than average. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3x = -\sqrt{3}/2\) | M1 | Accept \(-0.866\) at this stage |
| \(x = \dfrac{-\sqrt{3}}{6}\) oe | A1 [2] | Or \(\dfrac{-3}{6\sqrt{3}}\) or \(\dfrac{-1}{2\sqrt{3}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((2\cos\theta - 1)(\sin\theta - 1) = 0\) | M1 | Reasonable attempt to factorise and solve |
| \(\cos\theta = 1/2\) or \(\sin\theta = 1\) | A1 | Award B1B1 www |
| \(\theta = \pi/3\) or \(\pi/2\) | A1A1 [4] | Allow 1.05, 1.57. SCA1 for both 60°, 90° |
## Question 4:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3x = -\sqrt{3}/2$ | M1 | Accept $-0.866$ at this stage |
| $x = \dfrac{-\sqrt{3}}{6}$ oe | A1 [2] | Or $\dfrac{-3}{6\sqrt{3}}$ or $\dfrac{-1}{2\sqrt{3}}$ |
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(2\cos\theta - 1)(\sin\theta - 1) = 0$ | M1 | Reasonable attempt to factorise and solve |
| $\cos\theta = 1/2$ or $\sin\theta = 1$ | A1 | Award B1B1 www |
| $\theta = \pi/3$ or $\pi/2$ | A1A1 [4] | Allow 1.05, 1.57. SCA1 for both 60°, 90° |
---4
\begin{enumerate}[label=(\alph*)]
\item Solve the equation $\sin ^ { - 1 } ( 3 x ) = - \frac { 1 } { 3 } \pi$, giving the solution in an exact form.
\item Solve, by factorising, the equation $2 \cos \theta \sin \theta - 2 \cos \theta - \sin \theta + 1 = 0$ for $0 \leqslant \theta \leqslant \pi$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2016 Q4 [6]}}