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 \(\cot \frac{\pi }{8} = \tan \frac{8}{\pi }\).