| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| 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 routine application of R cos(x+α) = R cos α cos x - R sin α sin x, followed by straightforward equation solving. The multi-part structure and numerical accuracy requirements add slight complexity, but all steps follow textbook procedures with no novel insight needed, making it slightly easier than average. |
| 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/Working | Mark | Guidance |
| \(R = \sqrt{2^2 + 5^2} = \sqrt{29}\) | B1 | |
| \(\tan\alpha = \frac{5}{2}\), so \(\alpha = \arctan\left(\frac{5}{2}\right)\) | M1 | |
| \(\alpha = 1.19\) radians | A1 | 3 s.f. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Maximum when \(\cos(x + \alpha) = 1\), i.e. \(x + \alpha = 0\) or \(2\pi\) | M1 | |
| \(x = 2\pi - 1.19 = 5.09\) | A1 | 3 s.f. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sqrt{29}\cos(x+\alpha) = -1\), so \(\cos(x+\alpha) = \frac{-1}{\sqrt{29}}\) | M1 | |
| \(x + \alpha = \arccos\left(\frac{-1}{\sqrt{29}}\right) = 1.758...\) or \(2\pi - 1.758...\) | ||
| \(x = 1.758 - 1.190 = 0.568\) or \(x = 4.525 - 1.190 = ...\) giving \(x = 0.568\) and \(x = 3.71\) | A1 A1 | Both values, 3 s.f. |
# Question 2:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = \sqrt{2^2 + 5^2} = \sqrt{29}$ | B1 | |
| $\tan\alpha = \frac{5}{2}$, so $\alpha = \arctan\left(\frac{5}{2}\right)$ | M1 | |
| $\alpha = 1.19$ radians | A1 | 3 s.f. |
## Part (b)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Maximum when $\cos(x + \alpha) = 1$, i.e. $x + \alpha = 0$ or $2\pi$ | M1 | |
| $x = 2\pi - 1.19 = 5.09$ | A1 | 3 s.f. |
## Part (b)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sqrt{29}\cos(x+\alpha) = -1$, so $\cos(x+\alpha) = \frac{-1}{\sqrt{29}}$ | M1 | |
| $x + \alpha = \arccos\left(\frac{-1}{\sqrt{29}}\right) = 1.758...$ or $2\pi - 1.758...$ | | |
| $x = 1.758 - 1.190 = 0.568$ or $x = 4.525 - 1.190 = ...$ giving $x = 0.568$ and $x = 3.71$ | A1 A1 | Both values, 3 s.f. |
---2
\begin{enumerate}[label=(\alph*)]
\item Express $2 \cos x - 5 \sin x$ in the form $R \cos ( x + \alpha )$, where $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$, giving your value of $\alpha$, in radians, to three significant figures.
\item \begin{enumerate}[label=(\roman*)]
\item Hence find the value of $x$ in the interval $0 < x < 2 \pi$ for which $2 \cos x - 5 \sin x$ has its maximum value. Give your value of $x$ to three significant figures.
\item Use your answer to part (a) to solve the equation $2 \cos x - 5 \sin x + 1 = 0$ in the interval $0 < x < 2 \pi$, giving your solutions to three significant figures.\\[0pt]
[3 marks]
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C4 2015 Q2 [8]}}