| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Solve equation with reciprocal functions |
| Difficulty | Standard +0.3 This is a standard reciprocal trig equation requiring conversion to sin θ (using tan = sin/cos, cot = cos/sin, sec = 1/cos), then multiplying through by sin θ cos θ to clear fractions, yielding a quadratic in sin θ. The algebraic manipulation is routine for P2 level, and solving the resulting quadratic plus finding angles in the given range are well-practiced techniques. Slightly above average difficulty due to multiple steps, but follows a predictable method. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt to express left hand side of the equation in terms of \(\sin\theta\) and \(\cos\theta\) | M1 | With at least two of the terms correct and no missing \(\theta\)s. Condone use of \(x\) instead of \(\theta\) |
| Obtain \(7\sin^2\theta + 4\cos^2\theta - 13\sin\theta [= 0]\) | A1 | SOI, OE |
| Obtain \(3\sin^2\theta - 13\sin\theta + 4 = 0\) | A1 | Allow if missing \(\theta\)s are recovered. SC: Allow full marks for \(3\sin\theta - 13 + \dfrac{4}{\sin\theta} = 0\). Must be in terms of \(\theta\) for final A mark |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt solution of 3-term quadratic equation for \(\sin\theta\) | M1 | |
| Obtain \(\sin\theta = \frac{1}{3}\) and hence \(19.5°\) | A1 | or greater accuracy |
| Obtain second value \(160.5°\) | A1 | or greater accuracy; no other values within given range; FT on \(180° - \textit{their}\ 19.5°\) |
## Question 2:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt to express left hand side of the equation in terms of $\sin\theta$ and $\cos\theta$ | M1 | With at least two of the terms correct and no missing $\theta$s. Condone use of $x$ instead of $\theta$ |
| Obtain $7\sin^2\theta + 4\cos^2\theta - 13\sin\theta [= 0]$ | A1 | SOI, OE |
| Obtain $3\sin^2\theta - 13\sin\theta + 4 = 0$ | A1 | Allow if missing $\theta$s are recovered. SC: Allow full marks for $3\sin\theta - 13 + \dfrac{4}{\sin\theta} = 0$. Must be in terms of $\theta$ for final A mark |
| **Total** | **3** | |
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt solution of 3-term quadratic equation for $\sin\theta$ | M1 | |
| Obtain $\sin\theta = \frac{1}{3}$ and hence $19.5°$ | A1 | or greater accuracy |
| Obtain second value $160.5°$ | A1 | or greater accuracy; no other values within given range; FT on $180° - \textit{their}\ 19.5°$ |
---2
\begin{enumerate}[label=(\alph*)]
\item Express the equation $7 \tan \theta + 4 \cot \theta - 13 \sec \theta = 0$ in terms of $\sin \theta$ only.
\item Hence solve the equation $7 \tan \theta + 4 \cot \theta - 13 \sec \theta = 0$ for $0 ^ { \circ } < \theta < 360 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2022 Q2 [6]}}