| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2019 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Express double angle or product |
| Difficulty | Standard +0.3 This is a standard harmonic form question with routine techniques: converting to R sin(θ-α) form using Pythagorean identity, solving a transformed equation, applying double angle identities, and finding a maximum value. All steps follow textbook procedures with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Notes |
| \(R = \sqrt{53}\) | B1 | cao |
| \(\tan \alpha = \frac{2}{7} \Rightarrow \alpha = \ldots\) using \(\tan\alpha = \pm\frac{2}{7}\) or \(\tan\alpha = \pm\frac{7}{2}\) or \(\sin\alpha = \pm\frac{2}{\sqrt{53}}\) or \(\sin\alpha = \pm\frac{7}{\sqrt{53}}\) or \(\cos\alpha = \pm\frac{7}{\sqrt{53}}\) or \(\cos\alpha = \pm\frac{2}{\sqrt{53}}\) | M1 | Uses one of these equations to find a value for \(\alpha\) |
| \(\alpha = 15.95°\) | A1 | Awrt \(15.95°\) (Allow awrt \(0.28\) rad) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Notes |
| \(\sqrt{53}\sin(2\theta - 15.95°) = 4 \Rightarrow \sin(2\theta - 15.95°) = \frac{4}{\sqrt{53}}\) (0.549); attempts to use part (a) \(\sqrt{53}\sin(2\theta - \text{"}{15.95°}\text{"}) = 4\) and proceeds to \(\sin(2\theta \pm \text{"}{15.95°}\text{"}) = K\), \( | K | < 1\) |
| \(2\theta - 15.95° = 33.3287\ldots \Rightarrow \theta = 24.6°\) | A1 | Awrt \(24.6°\) (Allow awrt \(0.43\) rad) |
| \(2\theta - 15.95° = 180° - 33.287\ldots \Rightarrow \theta = \ldots\); e.g. \(2\theta_2 \mp 15.95° = 180° - \text{'}33.3287\ldots\text{'} \Rightarrow \theta_2 = \frac{180° - \text{'}33.3287\ldots\text{'} \pm \text{'}15.95°\text{'}}{2}\) (may be implied by their \(\theta_2\)); dependent upon having scored the previous M; do not allow mixing of radians and degrees — if working in radians must use \(\pi\) not 180 | dM1 | |
| \(\theta = 81.3°\) | A1 | Awrt \(81.3°\) only; ignore extra answers outside range but deduct the final A for extra answers in range |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Notes |
| \(28\sin\theta\cos\theta = a\sin2\theta \Rightarrow a = 14\) | B1 | \(a = 14\) |
| \(8\sin^2\theta = b\!\left(\pm 1 \pm 2\sin^2\theta\right) + c\) or \(8\sin^2\theta = 8\!\left(\frac{1}{2}(\pm 1 \pm \cos2\theta)\right)\); or \(8\sin^2\theta = 4\sin^2\theta + 4\sin^2\theta = 4\sin^2\theta + 4(1-\cos^2\theta) = \pm 4\cos2\theta \pm 4\); attempts to use a \(\cos2\theta\) identity e.g. \(\cos2\theta = \pm 1 \pm 2\sin^2\theta\) or \(\sin^2\theta = \frac{1}{2}(\pm 1 \pm \cos2\theta)\) at some point in their working and applies it to the given expression | M1 | |
| \(b = -4,\ c = 4\) or \(\ldots - 4\cos2\theta + 4\) | A1 | Correct values or correct expression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Notes |
| \(\left(28\sin\theta\cos\theta + 8\sin^2\theta\right)_{\max} = 2\text{'}\sqrt{53}\text{'} + \text{'}4\text{'}\) | M1 | Maximum \(= 2 \times \text{their}\,\sqrt{53} + \text{their}\,c\); may be implied e.g. by their decimal answer; attempts to use calculus for the maximum should reach \(2R + c\) as above for M1 |
| \(2\sqrt{53} + 4\) | A1 | cao (must be exact, not decimals) |
# Question 1:
## Part (a)
| Answer/Working | Mark | Notes |
|---|---|---|
| $R = \sqrt{53}$ | B1 | cao |
| $\tan \alpha = \frac{2}{7} \Rightarrow \alpha = \ldots$ using $\tan\alpha = \pm\frac{2}{7}$ or $\tan\alpha = \pm\frac{7}{2}$ or $\sin\alpha = \pm\frac{2}{\sqrt{53}}$ or $\sin\alpha = \pm\frac{7}{\sqrt{53}}$ or $\cos\alpha = \pm\frac{7}{\sqrt{53}}$ or $\cos\alpha = \pm\frac{2}{\sqrt{53}}$ | M1 | Uses one of these equations to find a value for $\alpha$ |
| $\alpha = 15.95°$ | A1 | Awrt $15.95°$ (Allow awrt $0.28$ rad) |
## Part (b)
| Answer/Working | Mark | Notes |
|---|---|---|
| $\sqrt{53}\sin(2\theta - 15.95°) = 4 \Rightarrow \sin(2\theta - 15.95°) = \frac{4}{\sqrt{53}}$ (0.549); attempts to use part (a) $\sqrt{53}\sin(2\theta - \text{"}{15.95°}\text{"}) = 4$ and proceeds to $\sin(2\theta \pm \text{"}{15.95°}\text{"}) = K$, $|K| < 1$ | M1 | Allow letter $\alpha$ for "15.95" |
| $2\theta - 15.95° = 33.3287\ldots \Rightarrow \theta = 24.6°$ | A1 | Awrt $24.6°$ (Allow awrt $0.43$ rad) |
| $2\theta - 15.95° = 180° - 33.287\ldots \Rightarrow \theta = \ldots$; e.g. $2\theta_2 \mp 15.95° = 180° - \text{'}33.3287\ldots\text{'} \Rightarrow \theta_2 = \frac{180° - \text{'}33.3287\ldots\text{'} \pm \text{'}15.95°\text{'}}{2}$ (may be implied by their $\theta_2$); dependent upon having scored the previous M; **do not allow mixing of radians and degrees — if working in radians must use $\pi$ not 180** | dM1 | |
| $\theta = 81.3°$ | A1 | Awrt $81.3°$ **only**; ignore extra answers outside range but deduct the final A for extra answers in range |
## Part (c)
| Answer/Working | Mark | Notes |
|---|---|---|
| $28\sin\theta\cos\theta = a\sin2\theta \Rightarrow a = 14$ | B1 | $a = 14$ |
| $8\sin^2\theta = b\!\left(\pm 1 \pm 2\sin^2\theta\right) + c$ or $8\sin^2\theta = 8\!\left(\frac{1}{2}(\pm 1 \pm \cos2\theta)\right)$; or $8\sin^2\theta = 4\sin^2\theta + 4\sin^2\theta = 4\sin^2\theta + 4(1-\cos^2\theta) = \pm 4\cos2\theta \pm 4$; attempts to use a $\cos2\theta$ identity e.g. $\cos2\theta = \pm 1 \pm 2\sin^2\theta$ or $\sin^2\theta = \frac{1}{2}(\pm 1 \pm \cos2\theta)$ at some point in their working and applies it to the given expression | M1 | |
| $b = -4,\ c = 4$ or $\ldots - 4\cos2\theta + 4$ | A1 | Correct values or correct expression |
## Part (d)
| Answer/Working | Mark | Notes |
|---|---|---|
| $\left(28\sin\theta\cos\theta + 8\sin^2\theta\right)_{\max} = 2\text{'}\sqrt{53}\text{'} + \text{'}4\text{'}$ | M1 | Maximum $= 2 \times \text{their}\,\sqrt{53} + \text{their}\,c$; may be implied e.g. by their decimal answer; attempts to use calculus for the maximum should reach $2R + c$ as above for M1 |
| $2\sqrt{53} + 4$ | A1 | cao (must be exact, not decimals) |\begin{enumerate}
\item (a) Express $7 \sin 2 \theta - 2 \cos 2 \theta$ in the form $R \sin ( 2 \theta - \alpha )$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha < 90 ^ { \circ }$. Give the exact value of $R$ and give the value of $\alpha$ to 2 decimal places.\\
(b) Hence solve, for $0 \leqslant \theta < 90 ^ { \circ }$, the equation
\end{enumerate}
$$7 \sin 2 \theta - 2 \cos 2 \theta = 4$$
giving your answers in degrees to one decimal place.\\
(c) Express $28 \sin \theta \cos \theta + 8 \sin ^ { 2 } \theta$ in the form $a \sin 2 \theta + b \cos 2 \theta + c$, where $a$, $b$ and $c$ are constants to be found.\\
(d) Use your answers to part (a) and part (c) to deduce the exact maximum value of $28 \sin \theta \cos \theta + 8 \sin ^ { 2 } \theta$\\
\hfill \mbox{\textit{Edexcel C34 2019 Q1 [12]}}