| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Express and solve equation |
| Difficulty | Standard +0.3 This is a standard harmonic form question requiring the R cos(θ - α) transformation, solving a trigonometric equation using the result, and identifying the range. All techniques are routine C3 material with straightforward application of formulas (R = √(a² + b²), tan α = b/a) and no novel problem-solving required. Slightly above average difficulty only due to the double angle and multi-part structure. |
| 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 | Guidance |
| \(R = \sqrt{4^2+2^2} = \sqrt{20} = 2\sqrt{5}\) | B1 | Condone \(R = \pm\sqrt{20}\) |
| \(\alpha = \arctan\left(\frac{1}{2}\right) = 26.565°\ldots\) | M1 | For \(\alpha=\arctan\left(\pm\frac{1}{2}\right)\) or \(\arctan(\pm 2)\) leading to solution; condone \(\cos\alpha=4, \sin\alpha=2\) |
| \(\alpha =\) awrt \(26.57°\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sqrt{20}\cos(2\theta - 26.6) = 1 \Rightarrow \cos(2\theta - 26.57) = \frac{1}{\sqrt{20}}\) | M1 | Allow \(\cos(\theta \pm \text{their } 26.57) = \frac{1}{\text{their }R}\) |
| \((2\theta - 26.57) = \pm 77.1\ldots° \Rightarrow \theta = \ldots\) | dM1 | Dependent on M1; correct order: deal with "26.57" before "2"; condone subtracting 26.57 instead of adding |
| \(\theta =\) awrt \(51.8°\) | A1 | |
| \(2\theta - 26.57 = -77.1\ldots° \Rightarrow \theta = -\) awrt \(25.3°\) | ddM1A1 | For correct method for secondary value; either \(2\theta \pm 26.57 = -\beta\) or \(2\theta \pm 26.57 = 360 - \beta\); withhold if extra solutions in range |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(k < -\sqrt{20}\) | B1ft | Score for one correct end; accept decimals including \(\sqrt{20} \approx\) awrt 4.5; condone \(g(\theta)>\sqrt{20}\), \(y>\sqrt{20}\) or boundaries included |
| \(k > \sqrt{20}\) | B1ft | For both intervals in terms of \(k\); accept \( |
# Question 3 (part a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = \sqrt{4^2+2^2} = \sqrt{20} = 2\sqrt{5}$ | B1 | Condone $R = \pm\sqrt{20}$ |
| $\alpha = \arctan\left(\frac{1}{2}\right) = 26.565°\ldots$ | M1 | For $\alpha=\arctan\left(\pm\frac{1}{2}\right)$ or $\arctan(\pm 2)$ leading to solution; condone $\cos\alpha=4, \sin\alpha=2$ |
| $\alpha =$ awrt $26.57°$ | A1 | |
# Question 3 (part b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sqrt{20}\cos(2\theta - 26.6) = 1 \Rightarrow \cos(2\theta - 26.57) = \frac{1}{\sqrt{20}}$ | M1 | Allow $\cos(\theta \pm \text{their } 26.57) = \frac{1}{\text{their }R}$ |
| $(2\theta - 26.57) = \pm 77.1\ldots° \Rightarrow \theta = \ldots$ | dM1 | Dependent on M1; correct order: deal with "26.57" before "2"; condone subtracting 26.57 instead of adding |
| $\theta =$ awrt $51.8°$ | A1 | |
| $2\theta - 26.57 = -77.1\ldots° \Rightarrow \theta = -$ awrt $25.3°$ | ddM1A1 | For correct method for secondary value; either $2\theta \pm 26.57 = -\beta$ or $2\theta \pm 26.57 = 360 - \beta$; withhold if extra solutions in range |
# Question 3 (part c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k < -\sqrt{20}$ | B1ft | Score for one correct end; accept decimals including $\sqrt{20} \approx$ awrt 4.5; condone $g(\theta)>\sqrt{20}$, $y>\sqrt{20}$ or boundaries included |
| $k > \sqrt{20}$ | B1ft | For both intervals in terms of $k$; accept $|k|>\sqrt{20}$; condone $k>\sqrt{20}, k<-\sqrt{20}$ written as "and"; $-\sqrt{20}>k>\sqrt{20}$ is B1 B0 |
---3.
$$g ( \theta ) = 4 \cos 2 \theta + 2 \sin 2 \theta$$
Given that $\mathrm { g } ( \theta ) = R \cos ( 2 \theta - \alpha )$, where $R > 0$ and $0 < \alpha < 90 ^ { \circ }$,
\begin{enumerate}[label=(\alph*)]
\item find the exact value of $R$ and the value of $\alpha$ to 2 decimal places.
\item Hence solve, for $- 90 ^ { \circ } < \theta < 90 ^ { \circ }$,
$$4 \cos 2 \theta + 2 \sin 2 \theta = 1$$
giving your answers to one decimal place.
Given that $k$ is a constant and the equation $\mathrm { g } ( \theta ) = k$ has no solutions,
\item state the range of possible values of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2015 Q3 [10]}}