Use De Moivre’s theorem to show that \({z^n} + {z^{ - n}} = 2\cos n\theta ,{\text{ }}n \in {\mathbb{Z}^ + }\).

Expand \({\left( {z + {z^{ - 1}}} \right)^4}\).

Hence show that \({\cos ^4}\theta  = p\cos 4\theta  + q\cos 2\theta  + r\), where \(p,{\text{ }}q\) and \(r\) are constants to be determined.

Show that \({\cos ^6}\theta  = \frac{1}{{32}}\cos 6\theta  + \frac{3}{{16}}\cos 4\theta  + \frac{{15}}{{32}}\cos 2\theta  + \frac{5}{{16}}\).

Hence find the value of \(\int_0^{\frac{\pi }{2}} {{{\cos }^6}\theta {\text{d}}\theta } \).

S is rotated through \(2\pi \) radians about the x-axis. Find the value of the volume generated.

(i)     Write down an expression for the constant term in the expansion of \({\left( {z + {z^{ - 1}}} \right)^{2k}}\), \(k \in {\mathbb{Z}^ + }\).

(ii)     Hence determine an expression for \(\int_0^{\frac{\pi }{2}} {{{\cos }^{2k}}\theta {\text{d}}\theta } \) in terms of k.

\({z^n} + {z^{ - n}} = \cos n\theta  + i\sin n\theta  + \cos ( - n\theta ) + i\sin ( - n\theta )\)     M1

\( = \cos n\theta  + \cos n\theta  + i\sin n\theta  - i\sin n\theta \)     A1

\( = 2\cos n\theta \)     AG

[2 marks]

(b)     \({\left( {z + {z^{ - 1}}} \right)^4} = {z^4} + 4{z^3}\left( {\frac{1}{z}} \right) + 6{z^2}\left( {\frac{1}{{{z^2}}}} \right) + 4z\left( {\frac{1}{{{z^3}}}} \right) + \frac{1}{{{z^4}}}\)     A1

Note:     Accept \({\left( {z + {z^{ - 1}}} \right)^4} = 16{\cos ^4}\theta \).

[1 mark]

METHOD 1

\({\left( {z + {z^{ - 1}}} \right)^4} = \left( {{z^4} + \frac{1}{{{z^4}}}} \right) + 4\left( {{z^2} + \frac{1}{{{z^2}}}} \right) + 6\)     M1

\({(2\cos \theta )^4} = 2\cos 4\theta  + 8\cos 2\theta  + 6\)     A1A1

Note:     Award A1 for RHS, A1 for LHS, independent of the M1.

\({\cos ^4}\theta  = \frac{1}{8}\cos 4\theta  + \frac{1}{2}\cos 2\theta  + \frac{3}{8}\)     A1

\(\left( {{\text{or }}p = \frac{1}{8},{\text{ }}q = \frac{1}{2},{\text{ }}r = \frac{3}{8}} \right)\)

METHOD 2

\({\cos ^4}\theta  = {\left( {\frac{{\cos 2\theta  + 1}}{2}} \right)^2}\)     M1

\( = \frac{1}{4}({\cos ^2}2\theta  + 2\cos 2\theta  + 1)\)     A1

\( = \frac{1}{4}\left( {\frac{{\cos 4\theta  + 1}}{2} + 2\cos 2\theta  + 1} \right)\)     A1

\({\cos ^4}\theta  = \frac{1}{8}\cos 4\theta  + \frac{1}{2}\cos 2\theta  + \frac{3}{8}\)     A1

\(\left( {{\text{or }}p = \frac{1}{8},{\text{ }}q = \frac{1}{2},{\text{ }}r = \frac{3}{8}} \right)\)

[4 marks]

\({\left( {z + {z^{ - 1}}} \right)^6} = {z^6} + 6{z^5}\left( {\frac{1}{z}} \right) + 15{z^4}\left( {\frac{1}{{{z^2}}}} \right) + 20{z^3}\left( {\frac{1}{{{z^3}}}} \right) + 15{z^2}\left( {\frac{1}{{{z^4}}}} \right) + 6z\left( {\frac{1}{{{z^5}}}} \right) + \frac{1}{{{z^6}}}\)     M1

\({\left( {z + {z^{ - 1}}} \right)^6} = \left( {{z^6} + \frac{1}{{{z^6}}}} \right) + 6\left( {{z^4} + \frac{1}{{{z^4}}}} \right) + 15\left( {{z^2} + \frac{1}{{{z^2}}}} \right) + 20\)

\({(2\cos \theta )^6} = 2\cos 6\theta  + 12\cos 4\theta  + 30\cos 2\theta  + 20\)     A1A1

Note:     Award A1 for RHS, A1 for LHS, independent of the M1.

\({\cos ^6}\theta  = \frac{1}{{32}}\cos 6\theta  + \frac{3}{{16}}\cos 4\theta  + \frac{{15}}{{32}}\cos 2\theta  + \frac{5}{{16}}\)     AG

Note:     Accept a purely trigonometric solution as for (c).

[3 marks]

\(\int_0^{\frac{\pi }{2}} {{{\cos }^6}\theta {\text{d}}\theta  = \int_0^{\frac{\pi }{2}} {\left( {\frac{1}{{32}}\cos 6\theta  + \frac{3}{{16}}\cos 4\theta  + \frac{{15}}{{32}}\cos 2\theta  + \frac{5}{{16}}} \right){\text{d}}\theta } } \)

\( = \left[ {\frac{1}{{192}}\sin 6\theta  + \frac{3}{{64}}\sin 4\theta  + \frac{{15}}{{64}}\sin 2\theta  + \frac{5}{{16}}\theta } \right]_0^{\frac{\pi }{2}}\)     M1A1

\( = \frac{{5\pi }}{{32}}\)     A1

[3 marks]

\({\text{V}} = \pi \int_0^{\frac{\pi }{2}} {{{\sin }^2}x{{\cos }^4}x{\text{d}}x} \)     M1

\( = \pi \int_0^{\frac{\pi }{2}} {{{\cos }^4}x{\text{d}}x - \pi \int_0^{\frac{\pi }{2}} {{{\cos }^6}x{\text{d}}x} } \)     M1

\(\int_0^{\frac{\pi }{2}} {{{\cos }^4}x{\text{d}}x}  = \frac{{3\pi }}{{16}}\)     A1

\({\text{V}} = \frac{{3{\pi ^2}}}{{16}} - \frac{{5{\pi ^2}}}{{32}} = \frac{{{\pi ^2}}}{{32}}\)     A1

Note:     Follow through from an incorrect r in (c) provided the final answer is positive.

(i)     constant term = \(\left( \begin{array}{c}2k\\k\end{array} \right)\) \( = \frac{{(2k)!}}{{k!k!}} = \frac{{(2k)!}}{{{{(k!)}^2}}}{\text{ (accept }}C_k^{2k})\)     A1

(ii)     \({2^{2k}}\int_0^{\frac{\pi }{2}} {{{\cos }^{2k}}\theta {\text{d}}\theta  = \frac{{(2k)!\pi }}{{{{(k!)}^2}}}\frac{\pi }{2}} \)     A1

          \(\int_0^{\frac{\pi }{2}} {{{\cos }^{2k}}\theta {\text{d}}\theta  = \frac{{(2k)!\pi }}{{{2^{2k + 1}}{{(k!)}^2}}}} \) \(\left( {{\rm{or}}\frac{{\left( \begin{array}{c}2k\\k\end{array} \right)\pi }}{{{2^{2k + 1}}}}} \right)\)     A1

[3 marks]

Part a) has appeared several times before, though with it again being a ‘show that’ question, some candidates still need to be more aware of the need to show every step in their working, including the result that \(\sin ( - n\theta ) =  - \sin (n\theta )\).

Part b) was usually answered correctly.

Part c) was again often answered correctly, though some candidates often less successfully utilised a trig-only approach rather than taking note of part b).

Part d) was a good source of marks for those who kept with the spirit of using complex numbers for this type of question. Some limited attempts at trig-only solutions were seen, and correct solutions using this approach were extremely rare.

Part e) was well answered, though numerical slips were often common. A small number integrated \(\sin n\theta \) as \(n\cos n\theta \).

A large number of candidates did not realise the help that part e) inevitably provided for part f). Some correctly expressed the volume as \(\pi \int {{{\cos }^4}x{\text{d}}x - \pi \int {{{\cos }^6}x{\text{d}}x} } \) and thus gained the first 2 marks but were able to progress no further. Only a small number of able candidates were able to obtain the correct answer of \(\frac{{{\pi ^2}}}{{32}}\).

Part g) proved to be a challenge for the vast majority, though it was pleasing to see some of the highest scoring candidates gain all 3 marks.