| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Substitution t equals tan |
| Difficulty | Standard +0.3 This is a structured multi-part question with clear guidance (formula recall, guided substitution, then solving). The algebraic manipulation in part (ii) requires careful handling of trig identities and the quartic in part (iii) factors nicely. While it involves multiple techniques, the scaffolding makes it slightly easier than average for C3. |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\tan 2x = \frac{2\tan x}{1 - \tan^2 x}\) | B1 | For correct RHS stated |
| (ii) \(\frac{8t}{1 - t^2} + 3x \cdot \frac{1}{1 - t^2} \cdot (1 + t^2) = 0\) Hence \(8t^2 + 3(1 - t^2)(1 + t^2) = 0\) i.e. \(3t^4 - 8t^3 - 3 = 0\), as required | B1, B1, M1, A1 | For \(\cot x = \frac{1}{t}\) seen; For \(\sec^2 x = 1 + t^2\) seen; For complete substitution in terms of \(t\); For showing given equation correctly |
| (iii) \((3t^2 + 1)(t^2 - 3) = 0\) Hence \(t = \pm\sqrt{3}\) So \(x = \frac{1}{3}\pi, \frac{2}{3}\pi, \frac{4}{3}\pi, \frac{5}{3}\pi\) | M1, A1, A1, A1 | For factorising or other solution method; For \(t^2 = 3\) found correctly; For any two correct angles; For all four correct and no others |
**(i)** $\tan 2x = \frac{2\tan x}{1 - \tan^2 x}$ | B1 | For correct RHS stated
**(ii)** $\frac{8t}{1 - t^2} + 3x \cdot \frac{1}{1 - t^2} \cdot (1 + t^2) = 0$ Hence $8t^2 + 3(1 - t^2)(1 + t^2) = 0$ i.e. $3t^4 - 8t^3 - 3 = 0$, as required | B1, B1, M1, A1 | For $\cot x = \frac{1}{t}$ seen; For $\sec^2 x = 1 + t^2$ seen; For complete substitution in terms of $t$; For showing given equation correctly
**(iii)** $(3t^2 + 1)(t^2 - 3) = 0$ Hence $t = \pm\sqrt{3}$ So $x = \frac{1}{3}\pi, \frac{2}{3}\pi, \frac{4}{3}\pi, \frac{5}{3}\pi$ | M1, A1, A1, A1 | For factorising or other solution method; For $t^2 = 3$ found correctly; For any two correct angles; For all four correct and no others
**Total for Question 7: 9 marks**
---7 (i) Write down the formula for $\tan 2 x$ in terms of $\tan x$.\\
(ii) By letting $\tan x = t$, show that the equation
$$4 \tan 2 x + 3 \cot x \sec ^ { 2 } x = 0$$
becomes
$$3 t ^ { 4 } - 8 t ^ { 2 } - 3 = 0$$
(iii) Hence find all the solutions of the equation
$$4 \tan 2 x + 3 \cot x \sec ^ { 2 } x = 0$$
which lie in the interval $0 \leqslant x \leqslant 2 \pi$.
\hfill \mbox{\textit{OCR C3 Q7 [9]}}