Show that $\frac{{\sin 2\theta }}{{1 + \cos 2\theta }} = \tan \theta$ .

Hence find the value of $\cot \frac{\pi }{8}$ in the form $a + b\sqrt 2$ , where $a,b \in \mathbb{Z}$.

$\frac{{\sin 2\theta }}{{1 + \cos 2\theta }} = \frac{{2\sin \theta \cos \theta }}{{1 + 2{{\cos }^2}\theta  - 1}}$     M1

Note: Award M1 for use of double angle formulae.

$= \frac{{2\sin \theta \cos \theta }}{{2{{\cos }^2}\theta }}$     A1

$= \frac{{\sin \theta }}{{\cos \theta }}$

$= \tan \theta$     AG

[2 marks]

$\tan \frac{\pi }{8} = \frac{{\sin \frac{\pi }{4}}}{{1 + \cos \frac{\pi }{4}}}$     (M1)

$\cot \frac{\pi }{8} = \frac{{1 + \cos \frac{\pi }{4}}}{{\sin \frac{\pi }{4}}}$     M1

$= \frac{{1 + \frac{{\sqrt 2 }}{2}}}{{\frac{{\sqrt 2 }}{2}}}$

$= 1 + \sqrt 2$     A1

[3 marks]

The performance in this question was generally good with most candidates answering (a) well; (b) caused more difficulties, in particular the rationalization of the denominator. A number of misconceptions were identified, for example $\cot \frac{\pi }{8} = \tan \frac{8}{\pi }$.

The performance in this question was generally good with most candidates answering (a) well; (b) caused more difficulties, in particular the rationalization of the denominator. A number of misconceptions were identified, for example $\cot \frac{\pi }{8} = \tan \frac{8}{\pi }$.