| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2018 |
| Session | March |
| Marks | 11 |
| Topic | Addition & Double Angle Formulae |
| Type | Find exact trigonometric values |
| Difficulty | Standard +0.3 Part (i) is a standard application of the tan subtraction formula with common angles (π/3 - π/4), requiring algebraic manipulation but following a well-established method. Part (ii) involves expressing a linear combination of sin and cos in R-form and solving, which is a routine A-level technique, though the 3A requires careful angle work. Both parts are textbook-standard with no novel insight required, making this slightly easier than average. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan \frac{\pi}{12} = \tan\left(\frac{\pi}{3} - \frac{\pi}{4}\right)\) | M1 | Any correct use of double angle formula |
| \(= \frac{\sqrt{3}-1}{1+\sqrt{3}}\) oe | A1 | Any correct expression for \(t\) (or correct QE) |
| \(= \frac{\sqrt{3}-1}{1+\sqrt{3}} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}\) | M1 | Attempts rationalising (or solve their QE) |
| \(= \frac{4-2\sqrt{3}}{2}\) | This form seen (or both roots) | |
| \(= 2 - \sqrt{3}\) (AG) | A1 | and correct answer alone |
| [4] |
| Answer | Marks |
|---|---|
| \(\frac{\sqrt{3}}{2}\sin 3A - \frac{1}{2}\cos 3A = \frac{1}{4}\) | M1 |
| \(\sin(3A - 30°) = \frac{1}{4}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(A = 14.8°\) | M1, A1 | Use of \(\sin^{-1}\) both sides |
| Answer | Marks |
|---|---|
| \(A = 65.2\) (1 dp) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(A = 134.8°\) | M1, A1if | ft their \(14.8° + 120°\) |
| [7] |
## 6(i)
DR
$\tan \frac{\pi}{12} = \tan\left(\frac{\pi}{3} - \frac{\pi}{4}\right)$ | M1 | Any correct use of double angle formula
$= \frac{\sqrt{3}-1}{1+\sqrt{3}}$ oe | A1 | Any correct expression for $t$ (or correct QE)
$= \frac{\sqrt{3}-1}{1+\sqrt{3}} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}$ | M1 | Attempts rationalising (or solve their QE)
$= \frac{4-2\sqrt{3}}{2}$ | | This form seen (or both roots)
$= 2 - \sqrt{3}$ (AG) | A1 | and correct answer alone
| [4]
## 6(ii)
DR
$\frac{\sqrt{3}}{2}\sin 3A - \frac{1}{2}\cos 3A = \frac{1}{4}$ | M1 |
$\sin(3A - 30°) = \frac{1}{4}$ | A1 |
$3A - 30° = 14.5$
$A = 14.8°$ | M1, A1 | Use of $\sin^{-1}$ both sides
or $3A - 30° = 165.5$
$A = 65.2$ (1 dp) | B1 |
or $3A - 30° = (14.5 + 360)°$
$A = 134.8°$ | M1, A1if | ft their $14.8° + 120°$
| [7]
---6 In this question you must show detailed reasoning.\\
(i) Use the formula for $\tan ( A - B )$ to show that $\tan \frac { \pi } { 12 } = 2 - \sqrt { 3 }$.\\
(ii) Solve the equation $2 \sqrt { 3 } \sin 3 A - 2 \cos 3 A = 1$ for $0 ^ { \circ } \leqslant A < 180 ^ { \circ }$.
\hfill \mbox{\textit{OCR H240/02 2018 Q6 [11]}}