The complex number z is defined as \(z = \cos \theta + {\text{i}}\sin \theta \) .

(a)     State de Moivre’s theorem.

(b)     Show that \({z^n} - \frac{1}{{{z^n}}} = 2{\text{i}}\sin (n\theta )\) .

(c)     Use the binomial theorem to expand \({\left( {z - \frac{1}{z}} \right)^5}\) giving your answer in simplified form.

(d)     Hence show that \(16{\sin ^5}\theta = \sin 5\theta - 5\sin 3\theta + 10\sin \theta \) .

(e)     Check that your result in part (d) is true for \(\theta = \frac{\pi }{4}\) .

(f)     Find \(\int_0^{\frac{\pi }{2}} {{{\sin }^5}\theta {\text{d}}\theta } \) .

(g)     Hence, with reference to graphs of circular functions, find \(\int_0^{\frac{\pi }{2}} {{{\cos }^5}\theta {\text{d}}\theta } \) , explaining your reasoning.

(a)     any appropriate form, e.g. \({(\cos \theta + {\text{i}}\sin \theta )^n} = \cos (n\theta) + {\text{i}}\sin (n\theta)\)     A1

[1 mark]

 

(b)     \({z^n} = \cos n\theta + {\text{i}}\sin n\theta \)     A1

\(\frac{1}{{{z^n}}} = \cos ( - n\theta ) + {\text{i}}\sin ( - n\theta )\)     (M1)

\( = \cos n\theta  - {\text{i}}\sin (n\theta )\)     A1

therefore \({z^n} - \frac{1}{{{z^n}}} = 2{\text{i}}\sin (n\theta )\)     AG

[3 marks]

 

(c)     \({\left( {z - \frac{1}{z}} \right)^5} = {z^5} + \left( {\begin{array}{*{20}{c}}
  5 \\
  1
\end{array}} \right){z^4}\left( { - \frac{1}{z}} \right) + \left( {\begin{array}{*{20}{c}}
  5 \\
  2
\end{array}} \right){z^3}{\left( { - \frac{1}{z}} \right)^2} + \left( {\begin{array}{*{20}{c}}
  5 \\
  3
\end{array}} \right){z^2}{\left( { - \frac{1}{z}} \right)^3} + \left( {\begin{array}{*{20}{c}}
  5 \\
  4
\end{array}} \right)z{\left( { - \frac{1}{z}} \right)^4} + {\left( { - \frac{1}{z}} \right)^5}\)     (M1)(A1)

\( = {z^5} - 5{z^3} + 10z - \frac{{10}}{z} + \frac{5}{{{z^3}}} - \frac{1}{{{z^5}}}\)     A1

[3 marks]

 

(d)     \({\left( {z - \frac{1}{z}} \right)^5} = {z^5} - \frac{1}{{{z^5}}} - 5\left( {{z^3} - \frac{1}{{{z^3}}}} \right) + 10\left( {z - \frac{1}{z}} \right)\)     M1A1

\({(2{\text{i}}\sin \theta )^5} = 2{\text{i}}\sin 5\theta - 10{\text{i}}\sin 3\theta + 20{\text{i}}\sin \theta \)     M1A1

\(16{\sin ^5}\theta = \sin 5\theta - 5\sin 3\theta + 10\sin \theta \)     AG

[4 marks]

 

(e)     \(16{\sin ^5}\theta = \sin 5\theta - 5\sin 3\theta + 10\sin \theta \)

\({\text{LHS}} = 16{\left( {\sin \frac{\pi }{4}} \right)^5}\)

\( = 16{\left( {\frac{{\sqrt 2 }}{2}} \right)^5}\)

\( = 2\sqrt 2 \,\,\,\,\,\left( { = \frac{4}{{\sqrt 2 }}} \right)\)     A1

\({\text{RHS}} = \sin \left( {\frac{{5\pi }}{4}} \right) - 5\sin \left( {\frac{{3\pi }}{4}} \right) + 10\sin \left( {\frac{\pi }{4}} \right)\)

\( = - \frac{{\sqrt 2 }}{2} - 5\left( {\frac{{\sqrt 2 }}{2}} \right) + 10\left( {\frac{{\sqrt 2 }}{2}} \right)\)     M1A1

Note: Award M1 for attempted substitution.

 

\( = 2\sqrt 2 \,\,\,\,\,\left( { = \frac{4}{{\sqrt 2 }}} \right)\)     A1

hence this is true for \(\theta = \frac{\pi }{4}\)     AG

[4 marks]

 

(f)     \(\int_0^{\frac{\pi }{2}} {{{\sin }^5}\theta {\text{d}}\theta = \frac{1}{{16}}\int_0^{\frac{\pi }{2}} {(\sin 5\theta - 5\sin 3\theta + 10\sin \theta ){\text{d}}\theta } } \)     M1

\( = \frac{1}{{16}}\left[ { - \frac{{\cos 5\theta }}{5} + \frac{{5\cos 3\theta }}{3} - 10\cos \theta } \right]_0^{\frac{\pi }{2}}\)     A1

\( = \frac{1}{{16}}\left[ {0 - \left( { - \frac{1}{5} + \frac{5}{3} - 10} \right)} \right]\)     A1

\( = \frac{8}{{15}}\)     A1

[4 marks]

 

(g)     \(\int_0^{\frac{\pi }{2}} {{{\cos }^5}\theta {\text{d}}\theta = \frac{8}{{15}}} \) , with appropriate reference to symmetry and graphs.     A1R1R1

Note: Award first R1 for partially correct reasoning e.g. sketches of graphs of sin and cos.

Award second R1 for fully correct reasoning involving \({\sin ^5}\) and \({\cos ^5}\).

 

[3 marks]

Total [22 marks]

Many students in b) substituted for the second term (again not making the connection to part a)) on the LHS and multiplied by the conjugate, which some managed well but it is inefficient. The binomial expansion was done well even if students did not do the earlier part. The connection between d) and f) was missed by many which lead to some creative attempts at the integral. Very few attempted the last part and of those many attempted another integral, ignoring the hence, while others related to the graph of sin and cos but not to the particular graphs here.