CAIE Further Paper 2 2022 November — Question 5 10 marks

Exam BoardCAIE
ModuleFurther Paper 2 (Further Paper 2)
Year2022
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeExpress roots in trigonometric form
DifficultyChallenging +1.2 Part (a) is routine recall of fourth roots of unity. Part (b) is a standard de Moivre's theorem application to derive a multiple angle formula—common Further Maths exercise. Part (c) requires recognizing the connection between parts and substituting x = cos θ, then solving cos 4θ = ±3/4, which involves some problem-solving but follows a well-established pattern for this topic. Overall, this is a structured multi-part question slightly above average difficulty due to the synthesis required in part (c).
Spec4.02q De Moivre's theorem: multiple angle formulae4.02r nth roots: of complex numbers

5
  1. Write down the fourth roots of unity.
  2. Use de Moivre's theorem to show that $$\cos 4 \theta = 8 \cos ^ { 4 } \theta - 8 \cos ^ { 2 } \theta + 1$$
  3. Hence obtain the real roots of the equation $$16 \left( 8 x ^ { 4 } - 8 x ^ { 2 } + 1 \right) ^ { 4 } - 9 = 0$$ in the form \(\cos ( q \pi )\), where \(q\) is a rational number.

Question 5(a):
AnswerMarks Guidance
AnswerMark Guidance
\(\pm 1,\,\pm i\)B1 Accept exponential form \(e^0, e^{i\frac{\pi}{2}}, e^{i\pi}, e^{i\frac{3\pi}{2}}\) or trigonometric form. Four values specified
1
Question 5(b):
AnswerMarks Guidance
AnswerMark Guidance
\(\cos 4\theta=\text{Re}(\cos\theta+i\sin\theta)^4=\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta\)M1 A1 Expands and takes real part
\(\cos^4\theta-6\cos^2\theta(1-\cos^2\theta)+(1-2\cos^2\theta+\cos^4\theta)\)M1 Applies \(\sin^2\theta=1-\cos^2\theta\)
\(8\cos^4\theta-8\cos^2\theta+1\)A1 AG. Can award final A1 without previous A1 if penultimate line seen. SC: Applying \(\cos 2A=2\cos^2 A-1\) twice scores 1/4
4
Question 5(c):
AnswerMarks Guidance
AnswerMark Guidance
\((8x^4-8x^2+1)^4=\frac{9}{16}\), \(\quad x=\cos\theta\)M1 Applies identity given in (b)
\(\cos 4\theta=\pm\frac{1}{2}\sqrt{3}\)A1 Need \(\pm\) and exact value for A1
\(4\theta=\pm\frac{1}{6}\pi+2k\pi \quad 4\theta=\pm\frac{5}{6}\pi+2k\pi\)M1 Solves \(\cos 4\theta=\frac{1}{2}\sqrt{3}\) or \(\cos 4\theta=-\frac{1}{2}\sqrt{3}\)
\(\cos(\frac{1}{24}\pi)\)A1 Gives one correct solution
\(\cos(\frac{11}{24}\pi),\cos(\frac{13}{24}\pi),\cos(\frac{23}{24}\pi),\) \(\cos(\frac{5}{24}\pi),\cos(\frac{7}{24}\pi),\cos(\frac{17}{24}\pi),\cos(\frac{19}{24}\pi)\)A1 Gives other solutions. Accept \(\pm\cos(\frac{1}{24}\pi),\pm\cos(\frac{11}{24}\pi),\pm\cos(\frac{5}{24}\pi),\pm\cos(\frac{7}{24}\pi)\)
5
## Question 5(a):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\pm 1,\,\pm i$ | **B1** | Accept exponential form $e^0, e^{i\frac{\pi}{2}}, e^{i\pi}, e^{i\frac{3\pi}{2}}$ or trigonometric form. Four values specified |
| | **1** | |

## Question 5(b):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\cos 4\theta=\text{Re}(\cos\theta+i\sin\theta)^4=\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta$ | **M1 A1** | Expands and takes real part |
| $\cos^4\theta-6\cos^2\theta(1-\cos^2\theta)+(1-2\cos^2\theta+\cos^4\theta)$ | **M1** | Applies $\sin^2\theta=1-\cos^2\theta$ |
| $8\cos^4\theta-8\cos^2\theta+1$ | **A1** | AG. Can award final A1 without previous A1 if penultimate line seen. SC: Applying $\cos 2A=2\cos^2 A-1$ twice scores 1/4 |
| | **4** | |

## Question 5(c):

| Answer | Mark | Guidance |
|--------|------|----------|
| $(8x^4-8x^2+1)^4=\frac{9}{16}$, $\quad x=\cos\theta$ | **M1** | Applies identity given in (b) |
| $\cos 4\theta=\pm\frac{1}{2}\sqrt{3}$ | **A1** | Need $\pm$ and exact value for A1 |
| $4\theta=\pm\frac{1}{6}\pi+2k\pi \quad 4\theta=\pm\frac{5}{6}\pi+2k\pi$ | **M1** | Solves $\cos 4\theta=\frac{1}{2}\sqrt{3}$ or $\cos 4\theta=-\frac{1}{2}\sqrt{3}$ |
| $\cos(\frac{1}{24}\pi)$ | **A1** | Gives one correct solution |
| $\cos(\frac{11}{24}\pi),\cos(\frac{13}{24}\pi),\cos(\frac{23}{24}\pi),$ $\cos(\frac{5}{24}\pi),\cos(\frac{7}{24}\pi),\cos(\frac{17}{24}\pi),\cos(\frac{19}{24}\pi)$ | **A1** | Gives other solutions. Accept $\pm\cos(\frac{1}{24}\pi),\pm\cos(\frac{11}{24}\pi),\pm\cos(\frac{5}{24}\pi),\pm\cos(\frac{7}{24}\pi)$ |
| | **5** | |

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5
\begin{enumerate}[label=(\alph*)]
\item Write down the fourth roots of unity.
\item Use de Moivre's theorem to show that

$$\cos 4 \theta = 8 \cos ^ { 4 } \theta - 8 \cos ^ { 2 } \theta + 1$$
\item Hence obtain the real roots of the equation

$$16 \left( 8 x ^ { 4 } - 8 x ^ { 2 } + 1 \right) ^ { 4 } - 9 = 0$$

in the form $\cos ( q \pi )$, where $q$ is a rational number.
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 2 2022 Q5 [10]}}