| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2019 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Express and solve equation |
| Difficulty | Moderate -0.3 This is a standard two-part harmonic form question requiring routine application of the R sin(x + α) formula with R = √(6+1) = √7, followed by solving a straightforward equation. While it involves exact values and a double angle substitution, these are well-practiced techniques at this level with no novel problem-solving required. |
| Spec | 1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State \(R = \sqrt{7}\) | B1 | |
| Use correct trig formulae to find \(\alpha\) | M1 | e.g. \(\tan\alpha = \dfrac{1}{\sqrt{6}}\), \(\sin\alpha = \dfrac{1}{\sqrt{7}}\), or \(\cos\alpha = \dfrac{\sqrt{6}}{\sqrt{7}}\) |
| Obtain \(\alpha = 22.208°\) | A1 | ISW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Evaluate \(\sin^{-1}\!\left(\dfrac{2}{\sqrt{7}}\right)\) to at least 1 d.p. | B1FT | \(49.107°\) to 3 d.p. B1 can be implied by correct answer(s) later. The FT is on *their* \(R\). SC: allow B1 for correct alternative equation e.g. \(3\tan^2\theta - 2\sqrt{6}\tan\theta + 1 = 0\) |
| Use correct method to find a value of \(\theta\) in the interval | M1 | Must get to \(\theta\) |
| Obtain answer, e.g. \(13.4°\) | A1 | Accept correct over-specified answers: \(13.449\ldots\), \(54.3425\ldots\) |
| Obtain second answer, e.g. \(54.3°\) and no extras in the given interval | A1 | Ignore answers outside the given interval |
## Question 4(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| State $R = \sqrt{7}$ | B1 | |
| Use correct trig formulae to find $\alpha$ | M1 | e.g. $\tan\alpha = \dfrac{1}{\sqrt{6}}$, $\sin\alpha = \dfrac{1}{\sqrt{7}}$, or $\cos\alpha = \dfrac{\sqrt{6}}{\sqrt{7}}$ |
| Obtain $\alpha = 22.208°$ | A1 | ISW |
---
## Question 4(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Evaluate $\sin^{-1}\!\left(\dfrac{2}{\sqrt{7}}\right)$ to at least 1 d.p. | B1FT | $49.107°$ to 3 d.p. B1 can be implied by correct answer(s) later. The FT is on *their* $R$. SC: allow B1 for correct alternative equation e.g. $3\tan^2\theta - 2\sqrt{6}\tan\theta + 1 = 0$ |
| Use correct method to find a value of $\theta$ in the interval | M1 | Must get to $\theta$ |
| Obtain answer, e.g. $13.4°$ | A1 | Accept correct over-specified answers: $13.449\ldots$, $54.3425\ldots$ |
| Obtain second answer, e.g. $54.3°$ and no extras in the given interval | A1 | Ignore answers outside the given interval |
---4 (i) Express $( \sqrt { } 6 ) \sin x + \cos x$ in the form $R \sin ( x + \alpha )$, where $R > 0$ and $0 ^ { \circ } < \alpha < 90 ^ { \circ }$. State the exact value of $R$ and give $\alpha$ correct to 3 decimal places.\\
(ii) Hence solve the equation $( \sqrt { } 6 ) \sin 2 \theta + \cos 2 \theta = 2$, for $0 ^ { \circ } < \theta < 180 ^ { \circ }$.\\
\hfill \mbox{\textit{CAIE P3 2019 Q4 [7]}}