6. (a) (i) By writing \(3 \theta = ( 2 \theta + \theta )\), show that
$$\sin 3 \theta = 3 \sin \theta - 4 \sin ^ { 3 } \theta$$
(ii) Hence, or otherwise, for \(0 < \theta < \frac { \pi } { 3 }\), solve
$$8 \sin ^ { 3 } \theta - 6 \sin \theta + 1 = 0 .$$
Give your answers in terms of \(\pi\).
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(a)(i) \(\sin 3\theta = \sin(2\theta + \theta)\)
Answer Marks
Guidance
\(= \sin 2\theta \cos \theta + \cos 2\theta \sin \theta\) M1 A1
\(= 2\sin\theta\cos\theta\cos\theta + (1-2\sin^2\theta)\sin\theta\) M1
\(= 2\sin\theta(1-\sin^2\theta) + \sin\theta - 2\sin^3\theta\) M1
\(= 3\sin\theta - 4\sin^3\theta\) cso
A1
(a)(ii)
Answer Marks
Guidance
\(8\sin^3\theta - 6\sin\theta + 1 = 0\) Given
\(-2\sin 3\theta + 1 = 0\) M1 A1
\(\sin 3\theta = \frac{1}{2}\) M1
\(3\theta = \frac{\pi}{6}, \frac{5\pi}{6}\) A1 A1
(5)
\(\theta = \frac{\pi}{18}, \frac{5\pi}{18}\)
(b)
Answer Marks
Guidance
\(\sin 15° = \sin(60° - 45°) = \sin 60° \cos 45° - \cos 60° \sin 45°\) M1
\(= \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}} - \frac{1}{2} \times \frac{1}{\sqrt{2}}\) M1 A1
\(= \frac{1}{4}\sqrt{6} - \frac{1}{4}\sqrt{2} = \frac{1}{4}(\sqrt{6}-\sqrt{2})\) cso
A1
Alternatives to (b):
Answer Marks
Guidance
\(\sin 15° = \sin(45° - 30°) = \sin 45° \cos 30° - \cos 45° \sin 30°\) M1
\(= \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \times \frac{1}{2}\) M1 A1
\(= \frac{1}{4}\sqrt{6} - \frac{1}{4}\sqrt{2} = \frac{1}{4}(\sqrt{6}-\sqrt{2})\) cso
A1
Using \(\cos 2\theta = 1 - 2\sin^2\theta\), \(\cos 30° = 1 - 2\sin^2 15°\)
Answer Marks
Guidance
\(2\sin^2 15° = 1 - \cos 30° = 1 - \frac{\sqrt{3}}{2}\) M1 A1
\(\sin^2 15° = \frac{2-\sqrt{3}}{4}\) M1
\(\left(\frac{1}{4}(\sqrt{6}-\sqrt{2})\right)^2 = \frac{1}{16}(6 - 2\sqrt{12}) = \frac{2-\sqrt{3}}{4}\) M1
Hence \(\sin 15° = \frac{1}{4}(\sqrt{6}-\sqrt{2})\) cso
A1
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**(a)(i)**
$\sin 3\theta = \sin(2\theta + \theta)$
$= \sin 2\theta \cos \theta + \cos 2\theta \sin \theta$ | M1 A1
$= 2\sin\theta\cos\theta\cos\theta + (1-2\sin^2\theta)\sin\theta$ | M1
$= 2\sin\theta(1-\sin^2\theta) + \sin\theta - 2\sin^3\theta$ | M1
$= 3\sin\theta - 4\sin^3\theta$ | cso | A1 | (4)
**(a)(ii)**
$8\sin^3\theta - 6\sin\theta + 1 = 0$ | Given
$-2\sin 3\theta + 1 = 0$ | M1 A1
$\sin 3\theta = \frac{1}{2}$ | M1
$3\theta = \frac{\pi}{6}, \frac{5\pi}{6}$ | A1 A1 | (5)
$\theta = \frac{\pi}{18}, \frac{5\pi}{18}$
**(b)**
$\sin 15° = \sin(60° - 45°) = \sin 60° \cos 45° - \cos 60° \sin 45°$ | M1
$= \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}} - \frac{1}{2} \times \frac{1}{\sqrt{2}}$ | M1 A1
$= \frac{1}{4}\sqrt{6} - \frac{1}{4}\sqrt{2} = \frac{1}{4}(\sqrt{6}-\sqrt{2})$ | cso | A1 | (4) [13]
**Alternatives to (b):**
$\sin 15° = \sin(45° - 30°) = \sin 45° \cos 30° - \cos 45° \sin 30°$ | M1
$= \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \times \frac{1}{2}$ | M1 A1
$= \frac{1}{4}\sqrt{6} - \frac{1}{4}\sqrt{2} = \frac{1}{4}(\sqrt{6}-\sqrt{2})$ | cso | A1 | (4)
Using $\cos 2\theta = 1 - 2\sin^2\theta$, $\cos 30° = 1 - 2\sin^2 15°$
$2\sin^2 15° = 1 - \cos 30° = 1 - \frac{\sqrt{3}}{2}$ | M1 A1
$\sin^2 15° = \frac{2-\sqrt{3}}{4}$ | M1
$\left(\frac{1}{4}(\sqrt{6}-\sqrt{2})\right)^2 = \frac{1}{16}(6 - 2\sqrt{12}) = \frac{2-\sqrt{3}}{4}$ | M1
Hence $\sin 15° = \frac{1}{4}(\sqrt{6}-\sqrt{2})$ | cso | A1 | (4)
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6. (a) (i) By writing $3 \theta = ( 2 \theta + \theta )$, show that
$$\sin 3 \theta = 3 \sin \theta - 4 \sin ^ { 3 } \theta$$
(ii) Hence, or otherwise, for $0 < \theta < \frac { \pi } { 3 }$, solve
$$8 \sin ^ { 3 } \theta - 6 \sin \theta + 1 = 0 .$$
Give your answers in terms of $\pi$.\\
(b) Using $\sin ( \theta - \alpha ) = \sin \theta \cos \alpha - \cos \theta \sin \alpha$, or otherwise, show that
$$\sin 15 ^ { \circ } = \frac { 1 } { 4 } ( \sqrt { } 6 - \sqrt { } 2 )$$
\hfill \mbox{\textit{Edexcel C3 2009 Q6 [13]}}