Higher June 2023 Paper 1 Q25
25 Show that \(\quad \dfrac{4\sin 30^\circ - \tan 45^\circ}{2\cos 30^\circ} \quad\) can be written as \(\tan x\), where \(x\) is an acute angle. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\sin 30 = \dfrac{1}{2}\) or \(\tan 45 = 1\) or \(\cos 30 = \dfrac{\sqrt{3}}{2}\) | M1 | oe eg \(\tan 45 = \dfrac{\sqrt{2}}{\sqrt{2}}\) or \(4\sin 30 = 2\) or \(2\cos 30 = \sqrt{3}\) implied by position in the expression may be seen in a table |
| substitution of all three correct values | M1dep | eg \(\dfrac{4 \times \frac{1}{2} - 1}{2 \times \frac{\sqrt{3}}{2}}\) or \(\dfrac{2 - 1}{2 \times \frac{\sqrt{3}}{2}}\) or \(\dfrac{2 - 1}{\sqrt{3}}\) |
| \(\dfrac{1}{\sqrt{3}}\) or \(\dfrac{\sqrt{3}}{3}\) | M1dep | |
| \(\left(\dfrac{1}{\sqrt{3}}\right.\) or \(\left.\dfrac{\sqrt{3}}{3} =\right)\ \tan 30\) or \(x = 30\) with full working seen for M3 | A1 |
Additional guidance
Allow \(\sqrt{1}\) for 1 throughout
Reference to 30° being an acute angle is not required