6 Use de Moivre's theorem to show that
$$\cos 5 \theta \equiv \cos \theta \left( 16 \sin ^ { 4 } \theta - 12 \sin ^ { 2 } \theta + 1 \right)$$
By considering the equation \(\cos 5 \theta = 0\), show that the exact value of \(\sin ^ { 2 } \left( \frac { 1 } { 10 } \pi \right)\) is \(\frac { 3 - \sqrt { 5 } } { 8 }\).
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Question 6:
Answer Marks
Guidance
Working/Answer Marks
Guidance
\((c+is)^5=c^5+5c^4is+10c^3(is)^2+10c^2(is)^3+5(is)^4+(is)^5\) B1
\(\cos 5\theta = c^5-10c^3s^2+5cs^4\) M1A1
\(=c\!\left((1-s^2)^2-10s^2(1-s^2)+5s^4\right)\) M1
\(=c(1-2s^2+s^4-10s^2+10s^4+5s^4)=c(16s^4-12s^2+1)\) A1 (5)
AG
Alternative using \((z+z^{-1})^n=2\cos n\theta\):\((2\cos\theta)^5=2\cos 5\theta+10\cos 3\theta+20\cos\theta\) (M1)
\(\Rightarrow 16\cos^5\theta=\cos 5\theta+5\cos 3\theta+10\cos\theta\) (B1)
But \(\cos 3\theta=4\cos^3\theta-3\cos\theta\) (can be quoted) \(\Rightarrow\cos 5\theta=\cos\theta(16\cos^4\theta-20\cos^2\theta+5)\) (A1)
Uses \(\cos^2\theta=1-\sin^2\theta\) to obtain \(\cos 5\theta=c(16s^4-12s^2+1)\) (M1A1)
AG
\(\cos 5\theta=0\Rightarrow\theta=\frac{1}{10}\pi,\frac{3}{10}\pi,\frac{1}{2}\pi,\frac{7}{10}\pi,\frac{9}{10}\pi\) B1
\(s^2=\frac{12\pm\sqrt{144-64}}{32}=\frac{3\pm\sqrt{5}}{8}\) M1A1
Since \(\sin^2\!\left(\tfrac{1}{10}\pi\right)<\sin^2\!\left(\tfrac{3}{10}\pi\right)\), \(\;\sin^2\!\left(\tfrac{1}{10}\pi\right)=\frac{3-\sqrt{5}}{8}\) A1 (4)
AG — justification required for final mark
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## Question 6:
| Working/Answer | Marks | Guidance |
|---|---|---|
| $(c+is)^5=c^5+5c^4is+10c^3(is)^2+10c^2(is)^3+5(is)^4+(is)^5$ | B1 | |
| $\cos 5\theta = c^5-10c^3s^2+5cs^4$ | M1A1 | |
| $=c\!\left((1-s^2)^2-10s^2(1-s^2)+5s^4\right)$ | M1 | |
| $=c(1-2s^2+s^4-10s^2+10s^4+5s^4)=c(16s^4-12s^2+1)$ | A1 (5) | AG |
| **Alternative** using $(z+z^{-1})^n=2\cos n\theta$: | | |
| $(2\cos\theta)^5=2\cos 5\theta+10\cos 3\theta+20\cos\theta$ | (M1) | |
| $\Rightarrow 16\cos^5\theta=\cos 5\theta+5\cos 3\theta+10\cos\theta$ | (B1) | |
| But $\cos 3\theta=4\cos^3\theta-3\cos\theta$ (can be quoted) | | |
| $\Rightarrow\cos 5\theta=\cos\theta(16\cos^4\theta-20\cos^2\theta+5)$ | (A1) | |
| Uses $\cos^2\theta=1-\sin^2\theta$ to obtain $\cos 5\theta=c(16s^4-12s^2+1)$ | (M1A1) | AG |
| $\cos 5\theta=0\Rightarrow\theta=\frac{1}{10}\pi,\frac{3}{10}\pi,\frac{1}{2}\pi,\frac{7}{10}\pi,\frac{9}{10}\pi$ | B1 | |
| $s^2=\frac{12\pm\sqrt{144-64}}{32}=\frac{3\pm\sqrt{5}}{8}$ | M1A1 | |
| Since $\sin^2\!\left(\tfrac{1}{10}\pi\right)<\sin^2\!\left(\tfrac{3}{10}\pi\right)$, $\;\sin^2\!\left(\tfrac{1}{10}\pi\right)=\frac{3-\sqrt{5}}{8}$ | A1 (4) | AG — justification required for final mark |
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6 Use de Moivre's theorem to show that
$$\cos 5 \theta \equiv \cos \theta \left( 16 \sin ^ { 4 } \theta - 12 \sin ^ { 2 } \theta + 1 \right)$$
By considering the equation $\cos 5 \theta = 0$, show that the exact value of $\sin ^ { 2 } \left( \frac { 1 } { 10 } \pi \right)$ is $\frac { 3 - \sqrt { 5 } } { 8 }$.
\hfill \mbox{\textit{CAIE FP1 2014 Q6 [9]}}