Standard +0.3 This is a straightforward algebraic manipulation question requiring students to divide the given expressions, square the result, and rationalize nested surds. While the algebra with nested surds requires care, the method is direct (tan = sin/cos, then square and simplify) with no problem-solving insight needed. The 3-mark allocation confirms it's a routine 'show that' exercise, slightly above average difficulty due to the nested surd manipulation but well within standard AS-level technique.
It is given that \(\cos 15Β° = \frac{1}{2}\sqrt{2 + \sqrt{3}}\) and \(\sin 15Β° = \frac{1}{2}\sqrt{2 - \sqrt{3}}\)
Show that \(\tan^2 15Β°\) can be written in the form \(a + b\sqrt{3}\), where \(a\) and \(b\) are integers.
Fully justify your answer.
[3 marks]
Question 9:
9 | π πππ
Uses tanπ = correctly
πππ π
(PI) (OE) | AO1.1a | M1 | β(2ββ3)
tan 15Β° =
β(2+β3)
2ββ3 2ββ3
tan215Β° = Γ
2+β3 2ββ3
2
= (2β β3)
tan215Β° = 7 β 4β3
Squares and multiplies by
conjugate
appropriate or vice versa
conjugate | AO1.1b | A1
Obtains correct values of a and b
(a = 7, b = β 4)
If no explicit evidence of the use of
conjugate
seen award max 1/3
conjugate
(M1 A0 R0) | AO2.1 | R1
Total | 3
Q | Marking Instructions | AO | Marks | Typical Solution
It is given that $\cos 15Β° = \frac{1}{2}\sqrt{2 + \sqrt{3}}$ and $\sin 15Β° = \frac{1}{2}\sqrt{2 - \sqrt{3}}$
Show that $\tan^2 15Β°$ can be written in the form $a + b\sqrt{3}$, where $a$ and $b$ are integers.
Fully justify your answer.
[3 marks]
\hfill \mbox{\textit{AQA AS Paper 2 2018 Q9 [3]}}