Moderate -0.8 This is a straightforward application of tan = sin/cos with given values, requiring only algebraic manipulation (rationalizing the denominator). The 4 marks reflect the working steps rather than conceptual difficulty. No problem-solving insight needed—purely procedural algebra that's easier than average A-level questions.
It is given that \(\sin 15° = \frac{\sqrt{6} - \sqrt{2}}{4}\) and \(\cos 15° = \frac{\sqrt{6} + \sqrt{2}}{4}\)
Use these two expressions to show that \(\tan 15° = 2 - \sqrt{3}\)
Fully justify your answer.
[4 marks]
Question 5:
5 | sin15°
Recalls tan 15° as
cos15°
OE PI by division of given surd
expressions ACF | 1.2 | B1 | sin15°
tan 15° =
cos15°
6− 2 6− 2
= ×
6+ 2 6− 2
6−2 12+2 8−4 3
=
6−2 4
= 2 – 3
Multiplies top and bottom by
conjugate of their denominator | 1.1a | M1
Expands either the denominator
or the numerator correctly
ACF | 1.1b | A1
Completes derivation of required
expression from correct
numerator and denominator
AG | 2.1 | R1
Question 5 Total | 4
Q | Marking instructions | AO | Marks | Typical solution
It is given that $\sin 15° = \frac{\sqrt{6} - \sqrt{2}}{4}$ and $\cos 15° = \frac{\sqrt{6} + \sqrt{2}}{4}$
Use these two expressions to show that $\tan 15° = 2 - \sqrt{3}$
Fully justify your answer.
[4 marks]
\hfill \mbox{\textit{AQA AS Paper 2 2023 Q5 [4]}}