| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Session | Specimen |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Find exact trigonometric values |
| Difficulty | Challenging +1.2 This question requires applying the tangent subtraction formula to find tan(15°) = tan(45° - 30°), then simplifying the resulting surd expression. While it involves multiple steps and careful algebraic manipulation of surds, the approach is fairly standard once students recognize 15° = 45° - 30°. The geometric setup in part (a) is straightforward for a regular 12-gon, and part (b) is a guided application of a formula from the specification. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.05l Double angle formulae: and compound angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Angle \(= 360 \div 24 = 15°\) | M1 | |
| Edge length \(= 2\tan 15°\) | ||
| Perimeter \(= 12 \times 2\tan 15° = 24\tan 15°\) | E1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\tan 15° = \tan(45°-30°)\) | B1 | |
| \(= \frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}} \left[= \frac{\sqrt{3}-1}{\sqrt{3}+1} = \frac{(\sqrt{3}-1)^2}{2}\right]\) | M1 | Exact values of \(\tan 45°\) and \(\tan 30°\) used |
| Alternative: \(\tan 15° = \tan(60°-45°) = \frac{\sqrt{3}-1}{1+\sqrt{3}} \left[= \frac{2\sqrt{3}-4}{-2}\right]\) | B1, M1 | Exact values of \(\tan 60°\) and \(\tan 15°\) used |
| Perimeter \(= 12 \times 2\tan 15° = 48 - 24\sqrt{3}\) | E1 | Correct completion; AG |
# Question 15:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Angle $= 360 \div 24 = 15°$ | M1 | |
| Edge length $= 2\tan 15°$ | | |
| Perimeter $= 12 \times 2\tan 15° = 24\tan 15°$ | E1 | AG |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\tan 15° = \tan(45°-30°)$ | B1 | |
| $= \frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}} \left[= \frac{\sqrt{3}-1}{\sqrt{3}+1} = \frac{(\sqrt{3}-1)^2}{2}\right]$ | M1 | Exact values of $\tan 45°$ and $\tan 30°$ used |
| **Alternative:** $\tan 15° = \tan(60°-45°) = \frac{\sqrt{3}-1}{1+\sqrt{3}} \left[= \frac{2\sqrt{3}-4}{-2}\right]$ | B1, M1 | Exact values of $\tan 60°$ and $\tan 15°$ used |
| Perimeter $= 12 \times 2\tan 15° = 48 - 24\sqrt{3}$ | E1 | Correct completion; AG |
---15 Fig. 15 shows a unit circle and the escribed regular polygon with 12 edges.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b4e10fd2-4144-4019-bf00-070f93a2b05d-11_839_876_356_269}
\captionsetup{labelformat=empty}
\caption{Fig. 15}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Show that the perimeter of the polygon is $24 \tan 15 ^ { \circ }$.
\item Using the formula for $\tan ( \theta - \phi )$ show that the perimeter of the polygon is $48 - 24 \sqrt { 3 }$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 Q15 [5]}}