| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Prove identity then solve equation |
| Difficulty | Moderate -0.3 This is a structured proof question with clear signposting ('show that', 'hence') requiring standard trigonometric identities. Part (i) is routine manipulation from the standard cos(α+β) formula, part (ii) follows directly by substitution, and part (iii) is straightforward equation solving using the derived result. The multi-step nature adds some complexity, but each step is mechanical with no novel insight required, making it slightly easier than average. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks |
|---|---|
| 4 (i) | 50 |
Question 4:
--- 4 (i) ---
4 (i) | 50
45
40
35
30
egatnecreP
25
20
15
10
5
0
18 22 26 30 34 38 42 46
Age
--- 4 (ii) ---
4 (ii)
--- 4 (iii) ---
4 (iii)
5 (A)
5 (B)\begin{enumerate}[label=(\roman*)]
\item Show that $\cos(\alpha + \beta) = \frac{1 - \tan \alpha \tan \beta}{\sec \alpha \sec \beta}$. [3]
\item Hence show that $\cos 2\alpha = \frac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha}$. [2]
\item Hence or otherwise solve the equation $\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \frac{1}{2}$ for $0° \leqslant \theta \leqslant 180°$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C4 2014 Q4 [8]}}