| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Prove identity then solve equation only (no integral) |
| Difficulty | Standard +0.8 This is a substantial multi-part question requiring proof of a non-trivial identity using addition formulae, solving a trigonometric equation involving both the proven identity and sec², and finally proving a general result about roots. The identity proof requires careful algebraic manipulation of tan addition formulae, part (ii) needs substitution and solving a quadratic in tan θ, and part (iii) requires understanding of the range and behavior of the function. While the techniques are C3 standard, the combination of steps and the need for strategic thinking elevates this above routine exercises. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals1.05p Proof involving trig: functions and identities |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Attempt use of either of \(\tan(A + B)\) identities | M1 | |
| Substitute \(\tan 60° = \sqrt{3}\) or \(\tan^2 60° = 3\) | B1 | |
| Obtain \(\frac{\tan \theta + \sqrt{3}}{1 - \sqrt{3}\tan \theta} \cdot \frac{\tan \theta - \sqrt{3}}{1 + \sqrt{3}\tan \theta}\) | A1 | or equiv (perhaps with \(\tan 60°\) still involved) |
| Obtain \(\frac{\tan^2 \theta - 3}{1 - 3\tan^2 \theta}\) | A1 | 4 AG |
| (ii) Use \(\sec^2 \theta = 1 + \tan^2 \theta\) | B1 | |
| Attempt rearrangement and simplification of equation involving \(\tan^2 \theta\) | M1 | or equiv involving \(\sec \theta\) |
| Obtain \(\tan^4 \theta = \frac{1}{3}\) | A1 | or equiv \(\sec^2 \theta = 1.57735…\) |
| Obtain \(37.2\) | A1 | or greater accuracy |
| Obtain \(142.8\) | A1 | 5 or greater accuracy; and no others between 0 and 180 |
| (iii) Attempt rearrangement of \(\frac{\tan^2 \theta - 3}{1 - 3\tan^2 \theta} = k^2\) to form | ||
| \(\tan^2 \theta = \frac{f(k)}{g(k)}\) | M1 | |
| Obtain \(\tan^2 \theta = \frac{k^2 + 3}{1 + 3k^2}\) | A1 | |
| Observe that RHS is positive for all \(k\), giving one value in each quadrant | A1 | 3 or convincing equiv |
(i) Attempt use of either of $\tan(A + B)$ identities | M1 |
Substitute $\tan 60° = \sqrt{3}$ or $\tan^2 60° = 3$ | B1 |
Obtain $\frac{\tan \theta + \sqrt{3}}{1 - \sqrt{3}\tan \theta} \cdot \frac{\tan \theta - \sqrt{3}}{1 + \sqrt{3}\tan \theta}$ | A1 | or equiv (perhaps with $\tan 60°$ still involved)
Obtain $\frac{\tan^2 \theta - 3}{1 - 3\tan^2 \theta}$ | A1 | 4 AG
(ii) Use $\sec^2 \theta = 1 + \tan^2 \theta$ | B1 |
Attempt rearrangement and simplification of equation involving $\tan^2 \theta$ | M1 | or equiv involving $\sec \theta$
Obtain $\tan^4 \theta = \frac{1}{3}$ | A1 | or equiv $\sec^2 \theta = 1.57735…$
Obtain $37.2$ | A1 | or greater accuracy
Obtain $142.8$ | A1 | 5 or greater accuracy; and no others between 0 and 180
(iii) Attempt rearrangement of $\frac{\tan^2 \theta - 3}{1 - 3\tan^2 \theta} = k^2$ to form | |
$\tan^2 \theta = \frac{f(k)}{g(k)}$ | M1 |
Obtain $\tan^2 \theta = \frac{k^2 + 3}{1 + 3k^2}$ | A1 |
Observe that RHS is positive for all $k$, giving one value in each quadrant | A1 | 3 or convincing equiv
---9 (i) Prove the identity
$$\tan \left( \theta + 60 ^ { \circ } \right) \tan \left( \theta - 60 ^ { \circ } \right) \equiv \frac { \tan ^ { 2 } \theta - 3 } { 1 - 3 \tan ^ { 2 } \theta }$$
(ii) Solve, for $0 ^ { \circ } < \theta < 180 ^ { \circ }$, the equation
$$\tan \left( \theta + 60 ^ { \circ } \right) \tan \left( \theta - 60 ^ { \circ } \right) = 4 \sec ^ { 2 } \theta - 3 ,$$
giving your answers correct to the nearest $0.1 ^ { \circ }$.\\
(iii) Show that, for all values of the constant k , the equation
$$\tan \left( \theta + 60 ^ { \circ } \right) \tan \left( \theta - 60 ^ { \circ } \right) = \mathrm { K } ^ { 2 }$$
has two roots in the interval $0 ^ { \circ } < \theta < 180 ^ { \circ }$.
\hfill \mbox{\textit{OCR C3 2007 Q9 [12]}}