June 2023 Paper 1 Q6
6
(a) Show that the equation \(\sin\left(x + \frac{1}{6}\pi\right) = \cos\left(x - \frac{1}{4}\pi\right)\) can be written in the form
\(\tan x = \dfrac{\sqrt{2}-1}{\sqrt{3}-\sqrt{2}}\). [4]
\(\tan x = \dfrac{\sqrt{2}-1}{\sqrt{3}-\sqrt{2}}\). [4]
(b) Hence solve the equation \(\sin\left(x + \frac{1}{6}\pi\right) = \cos\left(x - \frac{1}{4}\pi\right)\) for \(0 \leqslant x \leqslant 2\pi\). [1]
| Scheme | Marks | AO |
|---|---|---|
| \(\sin x\cos\dfrac{\pi}{6} + \cos x\sin\dfrac{\pi}{6}\) \(\quad = \cos x\cos\dfrac{\pi}{4} + \sin x\sin\dfrac{\pi}{4}\) | M1 | 2.1 |
| \(\dfrac{\sqrt{3}}{2}\sin x + \dfrac{1}{2}\cos x = \dfrac{\sqrt{2}}{2}\cos x + \dfrac{\sqrt{2}}{2}\sin x\) | M1 | 2.1 |
| \(\sin x\left(\dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{2}}{2}\right) = \cos x\left(\dfrac{\sqrt{2}}{2} - \dfrac{1}{2}\right)\) | M1 | 2.1 |
| \(\tan x = \dfrac{\sqrt{2}-1}{\sqrt{3}-\sqrt{2}}\) | E1 | 2.1 |
| [4] |
Notes
M1: Using a compound angle formula at least once
M1: Uses exact values for at least 2 trigonometric terms
M1: Collecting terms and factorising
E1: AG Complete argument with proper use of brackets where necessary
| Scheme | Marks | AO |
|---|---|---|
| \(x = \dfrac{7\pi}{24},\ \dfrac{31\pi}{24}\) | B1 | 1.1b |
| [1] |
Notes
B1: Allow for both values without working and no others in the range \(0 \leqslant x \leqslant 2\pi\). Allow decimal equivalents 0.916, 4.06 or better