| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Applied context modeling |
| Difficulty | Standard +0.3 This is a standard harmonic form question with straightforward geometric application. Part (a) is routine R-alpha conversion using standard formulas. Parts (b)-(d) involve basic trigonometric geometry with rectangles—the constraint equation comes from simple vertical distance considerations. While multi-part, each step follows predictable patterns with no novel insight required. Slightly easier than average due to the guided structure and standard techniques throughout. |
| Spec | 1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R = \sqrt{5} = 2.23606\ldots\) | B1 | Must be given in part (a); accept awrt 2.24 |
| \(\tan\alpha = \frac{1}{2}\) or \(\sin\alpha = \frac{1}{\sqrt{5}}\) or \(\cos\alpha = \frac{2}{\sqrt{5}}\) | M1 | Method mark may be implied by correct \(\alpha\) |
| \(\alpha = 26.56505\ldots°\) | A1 | Must be in part (a); accept awrt 26.57; also accept \(\sqrt{5}\sin(\theta + 26.57)\); answers in radians (0.46) are A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Way 1: Uses distance between two lines is 4 (or half distance is 2) with correct trigonometry; may state \(4\sin\theta + 2\cos\theta = 4\) or show sketch | M1 | |
| Need sketch and \(4\sin\theta + 2\cos\theta = 4\) and deduction that \(2\sin\theta + \cos\theta = 2\) or \(\cos\theta + 2\sin\theta = 2\) | A1* | Shows sketch with implication of two right-angled triangles and follows \(4\sin\theta + 2\cos\theta = 4\) by stating printed answer; no errors seen |
| Way 2: Uses diagonal of rectangle as hypotenuse; obtains \(\sqrt{20}\sin(\theta+\alpha) = 4\); so \(2\sin\theta + \cos\theta = 2\) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Way 1: Uses \(\sqrt{5}\sin(\theta + 26.57) = 2\); \(\sin(\theta + "26.57") = \frac{2}{"\sqrt{5}"}\) | M1 | Uses part (a) to solve equation |
| \(\theta = \arcsin\!\left(\frac{2}{\text{their }\sqrt{5}}\right) - "26.57"\) | M1 | Operations undone in correct order with subtraction |
| \(\theta = 36.8699\ldots°\) | A1 | awrt 36.9; answer in radians (0.644) is A0 |
| Way 2: Squares both sides, uses trig identities, reaches \(\tan\theta = \frac{3}{4}\) or \(\sin\theta = \frac{3}{5}\) or \(\cos\theta = \frac{4}{5}\) or \(\sin 2\theta = \frac{24}{25}\) | M1 | |
| \(\theta = \arctan\frac{3}{4}\) or other correct inverse trig value | M1 | |
| awrt 36.9 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Way 1: \("x" = \frac{2}{\tan 36.9}\), where \(x\) is the non-overlapping length of rectangle | B1 | States \(x = \frac{2}{\tan\theta}\) |
| \(\{h + x = 4 \Rightarrow\}\; h + \frac{2}{\tan"36.9"} = 4\) | M1 | Writes equation \(h + \frac{2}{\tan\theta} = 4\); must be this expression or equivalent |
| \(h = 4 - \frac{2}{\tan 36.9} = 1.336\ldots\) or \(\frac{4}{3}\) or \(\underline{1.3}\) (2 s.f.) | A1 cao | Accept decimal rounding to 1.3 or exact answer \(\frac{4}{3}\) |
| Way 2: \("y" = \frac{4}{\sin\theta}\); \(\{h+y=8\Rightarrow\}\; h + \frac{4}{\sin"36.9"} = 8\); \(h = 8 - \frac{4}{\sin 36.9} = \frac{4}{3}\) or \(\underline{1.3}\) (2 s.f.) | B1, M1, A1 cao |
# Question 13:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = \sqrt{5} = 2.23606\ldots$ | B1 | Must be given in part (a); accept awrt 2.24 |
| $\tan\alpha = \frac{1}{2}$ or $\sin\alpha = \frac{1}{\sqrt{5}}$ or $\cos\alpha = \frac{2}{\sqrt{5}}$ | M1 | Method mark may be implied by correct $\alpha$ |
| $\alpha = 26.56505\ldots°$ | A1 | Must be in part (a); accept awrt 26.57; also accept $\sqrt{5}\sin(\theta + 26.57)$; answers in radians (0.46) are A0 |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| **Way 1:** Uses distance between two lines is 4 (or half distance is 2) with correct trigonometry; may state $4\sin\theta + 2\cos\theta = 4$ or show sketch | M1 | |
| Need sketch **and** $4\sin\theta + 2\cos\theta = 4$ **and** deduction that $2\sin\theta + \cos\theta = 2$ or $\cos\theta + 2\sin\theta = 2$ | A1* | Shows sketch with implication of two right-angled triangles and follows $4\sin\theta + 2\cos\theta = 4$ by stating printed answer; no errors seen |
| **Way 2:** Uses diagonal of rectangle as hypotenuse; obtains $\sqrt{20}\sin(\theta+\alpha) = 4$; so $2\sin\theta + \cos\theta = 2$ | M1, A1 | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| **Way 1:** Uses $\sqrt{5}\sin(\theta + 26.57) = 2$; $\sin(\theta + "26.57") = \frac{2}{"\sqrt{5}"}$ | M1 | Uses part (a) to solve equation |
| $\theta = \arcsin\!\left(\frac{2}{\text{their }\sqrt{5}}\right) - "26.57"$ | M1 | Operations undone in correct order with subtraction |
| $\theta = 36.8699\ldots°$ | A1 | awrt 36.9; answer in radians (0.644) is A0 |
| **Way 2:** Squares both sides, uses trig identities, reaches $\tan\theta = \frac{3}{4}$ or $\sin\theta = \frac{3}{5}$ or $\cos\theta = \frac{4}{5}$ or $\sin 2\theta = \frac{24}{25}$ | M1 | |
| $\theta = \arctan\frac{3}{4}$ or other correct inverse trig value | M1 | |
| awrt 36.9 | A1 | |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| **Way 1:** $"x" = \frac{2}{\tan 36.9}$, where $x$ is the non-overlapping length of rectangle | B1 | States $x = \frac{2}{\tan\theta}$ |
| $\{h + x = 4 \Rightarrow\}\; h + \frac{2}{\tan"36.9"} = 4$ | M1 | Writes equation $h + \frac{2}{\tan\theta} = 4$; must be this expression or equivalent |
| $h = 4 - \frac{2}{\tan 36.9} = 1.336\ldots$ or $\frac{4}{3}$ or $\underline{1.3}$ (2 s.f.) | A1 cao | Accept decimal rounding to 1.3 or exact answer $\frac{4}{3}$ |
| **Way 2:** $"y" = \frac{4}{\sin\theta}$; $\{h+y=8\Rightarrow\}\; h + \frac{4}{\sin"36.9"} = 8$; $h = 8 - \frac{4}{\sin 36.9} = \frac{4}{3}$ or $\underline{1.3}$ (2 s.f.) | B1, M1, A1 cao | |13. (a) Express $2 \sin \theta + \cos \theta$ in the form $R \sin ( \theta + \alpha )$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha < 90 ^ { \circ }$. Give your value of $\alpha$ to 2 decimal places.\\
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{423eb549-0873-4185-8faf-12dedafcd108-21_467_1365_870_301}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows the design for a logo that is to be displayed on the side of a large building. The logo consists of three rectangles, $C , D$ and $E$, each of which is in contact with two horizontal parallel lines $l _ { 1 }$ and $l _ { 2 }$. Rectangle $D$ touches rectangles $C$ and $E$ as shown in Figure 4.
Rectangles $C , D$ and $E$ each have length 4 m and width 2 m . The acute angle $\theta$ between the line $l _ { 2 }$ and the longer edge of each rectangle is shown in Figure 4.
Given that $l _ { 1 }$ and $l _ { 2 }$ are 4 m apart,\\
(b) show that
$$2 \sin \theta + \cos \theta = 2$$
Given also that $0 < \theta < 45 ^ { \circ }$,\\
(c) solve the equation
$$2 \sin \theta + \cos \theta = 2$$
giving the value of $\theta$ to 1 decimal place.
Rectangles $C$ and $D$ and rectangles $D$ and $E$ touch for a distance $h \mathrm {~m}$ as shown in Figure 4.
Using your answer to part (c), or otherwise,\\
(d) find the value of $h$, giving your answer to 2 significant figures.
\hfill \mbox{\textit{Edexcel C34 2014 Q13 [11]}}