| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Quadratic trigonometric equations |
| Type | Show then solve: sin²/cos² substitution |
| Difficulty | Standard +0.3 Part (a) is straightforward reading of amplitude and vertical shift from a graph. Part (b)(i) requires expanding brackets and using sin²θ + cos²θ = 1 to convert to a quadratic in cos θ—a standard technique. Part (b)(ii) involves solving a simple quadratic and finding angles in a given range. This is a routine multi-part question testing standard P1 techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.05f Trigonometric function graphs: symmetries and periodicities1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks |
|---|---|
| \(a = -2\), \(b = 3\) | B1B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(s + s^2 - sc + 2c + 2sc - 2c^2 = s + sc \rightarrow s^2 - 2c^2 + 2c = 0\) | B1 | Expansion of brackets must be correct |
| \(1 - \cos^2\theta - 2\cos^2\theta + 2\cos\theta = 0\) | M1 | Uses \(s^2 = 1 - c^2\) |
| \(3\cos^2\theta - 2\cos\theta - 1 = 0\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos\theta = 1\) or \(-\dfrac{1}{3}\) | B1 | |
| \(\theta = 0°\) or \(109.5°\) or \(-109.5°\) | B1B1B1 FT | FT for \(-their\ 109.5°\) |
## Question 7(a):
$a = -2$, $b = 3$ | B1B1 |
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## Question 7(b)(i):
$s + s^2 - sc + 2c + 2sc - 2c^2 = s + sc \rightarrow s^2 - 2c^2 + 2c = 0$ | B1 | Expansion of brackets must be correct
$1 - \cos^2\theta - 2\cos^2\theta + 2\cos\theta = 0$ | M1 | Uses $s^2 = 1 - c^2$
$3\cos^2\theta - 2\cos\theta - 1 = 0$ | A1 | AG
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## Question 7(b)(ii):
$\cos\theta = 1$ or $-\dfrac{1}{3}$ | B1 |
$\theta = 0°$ or $109.5°$ or $-109.5°$ | B1B1B1 FT | FT for $-their\ 109.5°$
---7
\begin{enumerate}[label=(\alph*)]
\item \\
\includegraphics[max width=\textwidth, alt={}, center]{5201a3d5-7733-4d10-9de5-0c2255e3ad60-12_499_568_267_826}
The diagram shows part of the graph of $y = a + b \sin x$. Find the values of the constants $a$ and $b$.
\item \begin{enumerate}[label=(\roman*)]
\item Show that the equation
$$( \sin \theta + 2 \cos \theta ) ( 1 + \sin \theta - \cos \theta ) = \sin \theta ( 1 + \cos \theta )$$
may be expressed as $3 \cos ^ { 2 } \theta - 2 \cos \theta - 1 = 0$.
\item Hence solve the equation
$$( \sin \theta + 2 \cos \theta ) ( 1 + \sin \theta - \cos \theta ) = \sin \theta ( 1 + \cos \theta )$$
for $- 180 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2017 Q7 [9]}}