Question 7 - A-Level Maths

Edexcel FP2 2011 June — Question 7 11 marks

Exam Board	Edexcel
Module	FP2 (Further Pure Mathematics 2)
Year	2011
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Complex numbers 2
Type	Solve equations using trigonometric identities
Difficulty	Challenging +1.2 This is a standard Further Maths FP2 question requiring de Moivre's theorem to derive a multiple angle formula (part a), then solving a trigonometric equation using given identities (part b). While it involves multiple steps and Further Maths content, it follows a well-established template with clear guidance. The algebraic manipulation is moderately involved but routine for FP2 students.
Spec	1.05o Trigonometric equations: solve in given intervals 4.02q De Moivre's theorem: multiple angle formulae

(a) Use de Moivre's theorem to show that

$$\sin 5 \theta = 16 \sin ^ { 5 } \theta - 20 \sin ^ { 3 } \theta + 5 \sin \theta$$ Hence, given also that $\sin 3 \theta = 3 \sin \theta - 4 \sin ^ { 3 } \theta$,
(b) find all the solutions of $$\sin 5 \theta = 5 \sin 3 \theta$$ in the interval $0 \leqslant \theta < 2 \pi$. Give your answers to 3 decimal places.

Show mark scheme Show mark scheme source

Part (a):

Answer	Marks	Guidance
$\sin 5\theta = \text{Im}(\cos\theta + i\sin\theta)^5$	B1
$5\cos^4\theta(i\sin\theta) + 10\cos^2\theta(i^3\sin^3\theta) + i^5\sin^5\theta$	M1
$= i(5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta)$	A1
$(\text{Im}(\cos\theta + i\sin\theta)^5) = 5\sin\theta(1-\sin^2\theta)^2 - 10\sin^3\theta(1-\sin^2\theta) + \sin^5\theta$	M1
$\sin 5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta$	A1cso	$(*)$
(5)

Part (b):

Answer	Marks	Guidance
$16\sin^5\theta - 20\sin^3\theta + 5\sin\theta = 5(3\sin\theta - 4\sin^3\theta)$	M1
$16\sin^5\theta - 10\sin\theta = 0$	M1
$\sin^4\theta = \frac{5}{8}$	A1	$\theta = 1.095$
Inclusion of solutions from $\sin\theta = -\sqrt{\frac{5}{8}}$	M1
Other solutions: $\theta = 2.046, 4.237, 5.188$	A1
$\sin\theta = 0 \Rightarrow \theta = 0, \theta = \pi$	B1	$(3.142)$
(6) 11	Award B1 if solution considers Imaginary parts and equates to $\sin 5\theta$. 1st M1 for correct attempt at expansion and collection of imaginary parts. 2nd M1 for substitution powers of $\cos\theta$. 1st M for substituting correct expressions. 2nd M for attempting to form equation. Imply 3rd M if 4.237 or 5.188 seen. Award for their negative root. Ignore $2\pi$ but 2nd A0 if other extra solutions given.

**Part (a):**

| $\sin 5\theta = \text{Im}(\cos\theta + i\sin\theta)^5$ | B1 | |
| $5\cos^4\theta(i\sin\theta) + 10\cos^2\theta(i^3\sin^3\theta) + i^5\sin^5\theta$ | M1 | |
| $= i(5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta)$ | A1 | |
| $(\text{Im}(\cos\theta + i\sin\theta)^5) = 5\sin\theta(1-\sin^2\theta)^2 - 10\sin^3\theta(1-\sin^2\theta) + \sin^5\theta$ | M1 | |
| $\sin 5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta$ | A1cso | $(*)$ |
| | (5) | |

**Part (b):**

| $16\sin^5\theta - 20\sin^3\theta + 5\sin\theta = 5(3\sin\theta - 4\sin^3\theta)$ | M1 | |
| $16\sin^5\theta - 10\sin\theta = 0$ | M1 | |
| $\sin^4\theta = \frac{5}{8}$ | A1 | $\theta = 1.095$ |
| Inclusion of solutions from $\sin\theta = -\sqrt{\frac{5}{8}}$ | M1 | |
| Other solutions: $\theta = 2.046, 4.237, 5.188$ | A1 | |
| $\sin\theta = 0 \Rightarrow \theta = 0, \theta = \pi$ | B1 | $(3.142)$ |
| | (6) 11 | Award B1 if solution considers Imaginary parts and equates to $\sin 5\theta$. 1st M1 for correct attempt at expansion and collection of imaginary parts. 2nd M1 for substitution powers of $\cos\theta$. 1st M for substituting correct expressions. 2nd M for attempting to form equation. Imply 3rd M if 4.237 or 5.188 seen. Award for their negative root. Ignore $2\pi$ but 2nd A0 if other extra solutions given. |

Show LaTeX source

\begin{enumerate}
  \item (a) Use de Moivre's theorem to show that
\end{enumerate}

$$\sin 5 \theta = 16 \sin ^ { 5 } \theta - 20 \sin ^ { 3 } \theta + 5 \sin \theta$$

Hence, given also that $\sin 3 \theta = 3 \sin \theta - 4 \sin ^ { 3 } \theta$,\\
(b) find all the solutions of

$$\sin 5 \theta = 5 \sin 3 \theta$$

in the interval $0 \leqslant \theta < 2 \pi$. Give your answers to 3 decimal places.\\

\hfill \mbox{\textit{Edexcel FP2 2011 Q7 [11]}}

This paper (8 questions)

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Q1 7 Q2 7 Q3 8 Q4 9 Q5 9 Q6 9 Q7 11 Q8 15

Answer	Marks	Guidance
\(\sin 5\theta = \text{Im}(\cos\theta + i\sin\theta)^5\)	B1
\(5\cos^4\theta(i\sin\theta) + 10\cos^2\theta(i^3\sin^3\theta) + i^5\sin^5\theta\)	M1
\(= i(5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta)\)	A1
\((\text{Im}(\cos\theta + i\sin\theta)^5) = 5\sin\theta(1-\sin^2\theta)^2 - 10\sin^3\theta(1-\sin^2\theta) + \sin^5\theta\)	M1
\(\sin 5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta\)	A1cso	\((*)\)
(5)

Answer	Marks	Guidance
\(16\sin^5\theta - 20\sin^3\theta + 5\sin\theta = 5(3\sin\theta - 4\sin^3\theta)\)	M1
\(16\sin^5\theta - 10\sin\theta = 0\)	M1
\(\sin^4\theta = \frac{5}{8}\)	A1	\(\theta = 1.095\)
Inclusion of solutions from \(\sin\theta = -\sqrt{\frac{5}{8}}\)	M1
Other solutions: \(\theta = 2.046, 4.237, 5.188\)	A1
\(\sin\theta = 0 \Rightarrow \theta = 0, \theta = \pi\)	B1	\((3.142)\)
(6) 11	Award B1 if solution considers Imaginary parts and equates to \(\sin 5\theta\). 1st M1 for correct attempt at expansion and collection of imaginary parts. 2nd M1 for substitution powers of \(\cos\theta\). 1st M for substituting correct expressions. 2nd M for attempting to form equation. Imply 3rd M if 4.237 or 5.188 seen. Award for their negative root. Ignore \(2\pi\) but 2nd A0 if other extra solutions given.