6. (a) Use de Moivre's theorem to show that \(\boldsymbol { \operatorname { s i n } } 5 \boldsymbol { \theta } = \boldsymbol { \operatorname { s i n } } \boldsymbol { \theta } \left( \mathbf { 1 6 } \mathbf { c o s } ^ { 4 } \boldsymbol { \theta } - \mathbf { 1 2 } \boldsymbol { \operatorname { c o s } } ^ { 2 } \boldsymbol { \theta } + \mathbf { 1 } \right)\).
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(a) In this solution \(\cos\theta = c\) and \(\sin\theta = s\)\(\cos 5\theta + i\sin 5\theta = (c + is)^5\) M1
\((= c^5 + 5c^4is + 10c^3(is)^2 + 10c^2(is)^3 + 5c(is)^4 + (is)^5)\) \(\sin 5\theta = 5c^4s - 10c^3s^3 + s^5\) M1 A1
\(= 5c^4s - 10c^2(1-c^2)s + (1-c^2)^2s\) \(s^2 = 1 - c^2\) M1
\(= s(16c^4 - 12c^2 + 1)\) A1
5 marks
(b) \(\sin\theta(16\cos^4\theta - 12\cos^2\theta + 1) + 2\cos^2\theta\sin\theta = 0\)M1
\(\sin\theta = 0 \Rightarrow \theta = 0\) B1
\(16c^4 - 10c^2 + 1 = (8c^2 - 1)(2c^2 - 1) = 0\) M1
\(c = \pm\frac{1}{2\sqrt{2}}, c = \pm\frac{1}{\sqrt{2}}\) any two A1
\(\theta \approx 1.21, 1.93; \theta = \frac{\pi}{4}, \frac{3\pi}{4}\) any two A1
all four accept awrt \(0.79, 1.21, 1.93, 2.36\) Ignore any solutions out of range A1
6 marks [11]
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**(a)** In this solution $\cos\theta = c$ and $\sin\theta = s$ | |
$\cos 5\theta + i\sin 5\theta = (c + is)^5$ | M1 |
$(= c^5 + 5c^4is + 10c^3(is)^2 + 10c^2(is)^3 + 5c(is)^4 + (is)^5)$ | |
$\sin 5\theta = 5c^4s - 10c^3s^3 + s^5$ | M1 A1 |
$= 5c^4s - 10c^2(1-c^2)s + (1-c^2)^2s$ $s^2 = 1 - c^2$ | M1 |
$= s(16c^4 - 12c^2 + 1)$ | A1 | 5 marks
**(b)** $\sin\theta(16\cos^4\theta - 12\cos^2\theta + 1) + 2\cos^2\theta\sin\theta = 0$ | M1 |
$\sin\theta = 0 \Rightarrow \theta = 0$ | B1 |
$16c^4 - 10c^2 + 1 = (8c^2 - 1)(2c^2 - 1) = 0$ | M1 |
$c = \pm\frac{1}{2\sqrt{2}}, c = \pm\frac{1}{\sqrt{2}}$ any two | A1 |
$\theta \approx 1.21, 1.93; \theta = \frac{\pi}{4}, \frac{3\pi}{4}$ any two | A1 |
all four accept awrt $0.79, 1.21, 1.93, 2.36$ Ignore any solutions out of range | A1 | 6 marks [11]
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6. (a) Use de Moivre's theorem to show that $\boldsymbol { \operatorname { s i n } } 5 \boldsymbol { \theta } = \boldsymbol { \operatorname { s i n } } \boldsymbol { \theta } \left( \mathbf { 1 6 } \mathbf { c o s } ^ { 4 } \boldsymbol { \theta } - \mathbf { 1 2 } \boldsymbol { \operatorname { c o s } } ^ { 2 } \boldsymbol { \theta } + \mathbf { 1 } \right)$.\\
(b) Hence, or otherwise, solve, for $0 \leq \theta < \pi$
$$\sin 5 \theta + \cos \theta \sin 2 \theta = 0$$
(6)(Total 11 marks)\\
\hfill \mbox{\textit{Edexcel FP2 2006 Q6 [11]}}