Pre-U Pre-U 9794/2 2015 June — Question 10 14 marks

Exam BoardPre-U
ModulePre-U 9794/2 (Pre-U Mathematics Paper 2)
Year2015
SessionJune
Marks14
TopicStandard trigonometric equations
TypeProve identity then solve
DifficultyChallenging +1.2 This is a multi-part question requiring standard trigonometric identities (compound angle formulas), solving trigonometric equations, and making a connection to cubic equations via substitution. While it has several steps and the final part requires recognizing the link between the trigonometric form and the cubic, all techniques are standard A-level material with clear signposting ('hence' directs the approach). The cubic connection is moderately sophisticated but well-scaffolded, placing this above average difficulty but not requiring exceptional insight.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals
10
  1. Show that \(\sin \left( 2 \theta + \frac { 1 } { 2 } \pi \right) = \cos 2 \theta\).
  2. Hence solve the equation \(\sin 3 \theta = \cos 2 \theta\) for \(0 \leqslant \theta \leqslant 2 \pi\).
  3. Show that \(\sin 3 \theta = 3 \sin \theta - 4 \sin ^ { 3 } \theta\). Hence, by writing \(\cos 2 \theta - \sin 3 \theta\) in terms of \(\sin \theta\), use your answer to part (ii) to determine the solutions of \(4 x ^ { 3 } - 2 x ^ { 2 } - 3 x + 1 = 0\).
Question 10:
(i)
\[\sin\left(2\theta+\frac{1}{2}\pi\right) = \sin 2\theta\cos\frac{1}{2}\pi + \sin\frac{1}{2}\pi\cos 2\theta\]
\(\cos\frac{1}{2}\pi=0\), \(\sin\frac{1}{2}\pi=1\) so \(\sin\left(2\theta+\frac{1}{2}\pi\right)=\cos 2\theta\)
- M1: Use correct expansion (these values must be explicit or implied in method for A1)
- A1: Obtain given answer convincingly
Also allow arguments by linear transformations.
Total for (i): [2]
(ii)
\(\sin\left(2\theta+\frac{1}{2}\pi\right)=\sin 3\theta\)
A: \(2\theta+\frac{1}{2}\pi=3\theta \Rightarrow \theta=\frac{1}{2}\pi\)
B: \(3\theta = \pi-\left(2\theta+\frac{1}{2}\pi\right) \Rightarrow \theta=\frac{1}{10}\pi\)
\(3\theta=\pi-\left(2\theta+\frac{1}{2}\pi\right)+4\pi \Rightarrow \theta=\frac{9}{10}\pi\)
\(3\theta=\pi-\left(2\theta+\frac{1}{2}\pi\right)+6\pi \Rightarrow \theta=\frac{13}{10}\pi\)
\(3\theta=\pi-\left(2\theta+\frac{1}{2}\pi\right)+8\pi \Rightarrow \theta=\frac{17}{10}\pi\)
- B1: Obtain \(\frac{1}{2}\pi\)
- M1: Attempt second solution using symmetry of sin curve oe
- A1: Obtain \(\frac{1}{10}\pi\)
- A1: Obtain \(\frac{9}{10}\pi\)
- A1: Obtain \(\frac{13}{10}\pi\)
- A1: Obtain \(\frac{17}{10}\pi\)
Accept decimal equivalents for each root. After B1M1A1 given, apply penalty of \(-1\) against final three A marks for each additional incorrect root.
Total for (ii): [6]
(iii)
\[\sin(2\theta+\theta)=\sin 2\theta\cos\theta+\cos 2\theta\sin\theta\]
\[=2\sin\theta\cos^2\theta+(1-2\sin^2\theta)\sin\theta\]
\[=2\sin\theta-2\sin^3\theta+\sin\theta-2\sin^3\theta = 3\sin\theta-4\sin^3\theta\]
\(\cos 2\theta - \sin 3\theta = 0\):
\((1-2\sin^2\theta)-(3\sin\theta-4\sin^3\theta)=0\)
\(4\sin^3\theta-2\sin^2\theta-3\sin\theta+1=0\)
Let \(x=\sin\theta\): \(x=0.309,\ 1,\ -0.809\) to 3sf
- M1*: Expand using \(\sin(2\theta+\theta)\) or use De Moivre's theorem
- M1d*: Attempt to get expression in terms of \(\sin\theta\) only
- A1: Obtain given answer convincingly
- M1: Attempt to rearrange to comparable format
- M1: Identify \(x=\sin\theta\) (could be implied) and attempt to use solution(s) from part (ii)
- A1: Obtain \(x=0.309,\ 1,\ -0.809\) (allow 2dp). Allow surd values of \(\frac{1}{4}\left(-1\pm\sqrt{5}\right)\)
Total for (iii): [6]
**Question 10:**

**(i)**
$$\sin\left(2\theta+\frac{1}{2}\pi\right) = \sin 2\theta\cos\frac{1}{2}\pi + \sin\frac{1}{2}\pi\cos 2\theta$$
$\cos\frac{1}{2}\pi=0$, $\sin\frac{1}{2}\pi=1$ so $\sin\left(2\theta+\frac{1}{2}\pi\right)=\cos 2\theta$

- **M1**: Use correct expansion (these values must be explicit or implied in method for A1)
- **A1**: Obtain given answer convincingly

Also allow arguments by linear transformations.

**Total for (i): [2]**

**(ii)**
$\sin\left(2\theta+\frac{1}{2}\pi\right)=\sin 3\theta$

A: $2\theta+\frac{1}{2}\pi=3\theta \Rightarrow \theta=\frac{1}{2}\pi$

B: $3\theta = \pi-\left(2\theta+\frac{1}{2}\pi\right) \Rightarrow \theta=\frac{1}{10}\pi$

$3\theta=\pi-\left(2\theta+\frac{1}{2}\pi\right)+4\pi \Rightarrow \theta=\frac{9}{10}\pi$

$3\theta=\pi-\left(2\theta+\frac{1}{2}\pi\right)+6\pi \Rightarrow \theta=\frac{13}{10}\pi$

$3\theta=\pi-\left(2\theta+\frac{1}{2}\pi\right)+8\pi \Rightarrow \theta=\frac{17}{10}\pi$

- **B1**: Obtain $\frac{1}{2}\pi$
- **M1**: Attempt second solution using symmetry of sin curve oe
- **A1**: Obtain $\frac{1}{10}\pi$
- **A1**: Obtain $\frac{9}{10}\pi$
- **A1**: Obtain $\frac{13}{10}\pi$
- **A1**: Obtain $\frac{17}{10}\pi$

Accept decimal equivalents for each root. After B1M1A1 given, apply penalty of $-1$ against final three A marks for each additional incorrect root.

**Total for (ii): [6]**

**(iii)**
$$\sin(2\theta+\theta)=\sin 2\theta\cos\theta+\cos 2\theta\sin\theta$$
$$=2\sin\theta\cos^2\theta+(1-2\sin^2\theta)\sin\theta$$
$$=2\sin\theta-2\sin^3\theta+\sin\theta-2\sin^3\theta = 3\sin\theta-4\sin^3\theta$$

$\cos 2\theta - \sin 3\theta = 0$:
$(1-2\sin^2\theta)-(3\sin\theta-4\sin^3\theta)=0$
$4\sin^3\theta-2\sin^2\theta-3\sin\theta+1=0$

Let $x=\sin\theta$: $x=0.309,\ 1,\ -0.809$ to 3sf

- **M1***: Expand using $\sin(2\theta+\theta)$ or use De Moivre's theorem
- **M1d***: Attempt to get expression in terms of $\sin\theta$ only
- **A1**: Obtain given answer convincingly
- **M1**: Attempt to rearrange to comparable format
- **M1**: Identify $x=\sin\theta$ (could be implied) and attempt to use solution(s) from part **(ii)**
- **A1**: Obtain $x=0.309,\ 1,\ -0.809$ (allow 2dp). Allow surd values of $\frac{1}{4}\left(-1\pm\sqrt{5}\right)$

**Total for (iii): [6]**
10 (i) Show that $\sin \left( 2 \theta + \frac { 1 } { 2 } \pi \right) = \cos 2 \theta$.\\
(ii) Hence solve the equation $\sin 3 \theta = \cos 2 \theta$ for $0 \leqslant \theta \leqslant 2 \pi$.\\
(iii) Show that $\sin 3 \theta = 3 \sin \theta - 4 \sin ^ { 3 } \theta$. Hence, by writing $\cos 2 \theta - \sin 3 \theta$ in terms of $\sin \theta$, use your answer to part (ii) to determine the solutions of $4 x ^ { 3 } - 2 x ^ { 2 } - 3 x + 1 = 0$.

\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2015 Q10 [14]}}
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