6 Use de Moivre's theorem to show that
$$\cos 4 \theta = 8 \cos ^ { 4 } \theta - 8 \cos ^ { 2 } \theta + 1$$
Without using a calculator, verify that \(\cos 4 \theta = - \cos 3 \theta\) for each of the values \(\theta = \frac { 1 } { 7 } \pi , \frac { 3 } { 7 } \pi , \frac { 5 } { 7 } \pi , \pi\).
Using the result \(\cos 3 \theta = 4 \cos ^ { 3 } \theta - 3 \cos \theta\), show that the roots of the equation
$$8 c ^ { 4 } + 4 c ^ { 3 } - 8 c ^ { 2 } - 3 c + 1 = 0$$
are \(\cos \frac { 1 } { 7 } \pi , \cos \frac { 3 } { 7 } \pi , \cos \frac { 5 } { 7 } \pi , - 1\).
Deduce that \(\cos \frac { 1 } { 7 } \pi + \cos \frac { 3 } { 7 } \pi + \cos \frac { 5 } { 7 } \pi = \frac { 1 } { 2 }\).
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Question 6:
Answer Marks
Guidance
Working/Answer Mark
Guidance
\((\cos\theta + i\sin\theta)^4 = \cos 4\theta + i\sin 4\theta\) \(= c^4 + 4c^3(is) + 6c^2(is)^2 + 4c(is)^3 + (is)^4\) M1
\(\cos 4\theta = c^4 - 6c^2(1-c^2)+(1-c^2)^2\) M1
\(= 8\cos^4\theta - 8\cos^2\theta + 1\) (AG) A1
Part total 3
\(\cos\frac{4}{7}\pi = -\cos\left(\pi - \frac{4}{7}\pi\right) = -\cos\frac{3}{7}\pi\); \(\cos\frac{12}{7}\pi = \cos\left(-\frac{2}{7}\pi\right) = -\cos\left(\pi+\frac{2}{7}\pi\right) = -\cos\frac{9}{7}\pi\) B1
Verifies any two cases
\(\cos\frac{20}{7}\pi = \cos\frac{6}{7}\pi = -\cos\frac{1}{7}\pi = -\cos\frac{15}{7}\pi\); \(\cos 4\pi = 1 = -(-1) = -\cos 3\pi\) B1
Verifies remaining two cases; part total 2
\(8\cos^4\theta - 8\cos^2\theta + 1 = -(4\cos^3\theta - 3\cos\theta)\) \(\Rightarrow 8c^4 + 4c^3 - 8c^2 - 3c + 1 = 0\) (*) M1
Shows roots of equation
\(\Rightarrow \cos\frac{1}{7}\pi,\ \cos\frac{3}{7}\pi,\ \cos\frac{5}{7}\pi,\ -1\) are the roots (AG) A1
Part total 2
Sum of roots of (*) are \(\frac{-4}{8} = -\frac{1}{2}\) \(\Rightarrow\) Result (AG) M1A1
States sum of roots; part total 2; Total [9]
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## Question 6:
| Working/Answer | Mark | Guidance |
|---|---|---|
| $(\cos\theta + i\sin\theta)^4 = \cos 4\theta + i\sin 4\theta$ | | |
| $= c^4 + 4c^3(is) + 6c^2(is)^2 + 4c(is)^3 + (is)^4$ | M1 | |
| $\cos 4\theta = c^4 - 6c^2(1-c^2)+(1-c^2)^2$ | M1 | |
| $= 8\cos^4\theta - 8\cos^2\theta + 1$ (AG) | A1 | Part total 3 |
| $\cos\frac{4}{7}\pi = -\cos\left(\pi - \frac{4}{7}\pi\right) = -\cos\frac{3}{7}\pi$; $\cos\frac{12}{7}\pi = \cos\left(-\frac{2}{7}\pi\right) = -\cos\left(\pi+\frac{2}{7}\pi\right) = -\cos\frac{9}{7}\pi$ | B1 | Verifies any two cases |
| $\cos\frac{20}{7}\pi = \cos\frac{6}{7}\pi = -\cos\frac{1}{7}\pi = -\cos\frac{15}{7}\pi$; $\cos 4\pi = 1 = -(-1) = -\cos 3\pi$ | B1 | Verifies remaining two cases; part total 2 |
| $8\cos^4\theta - 8\cos^2\theta + 1 = -(4\cos^3\theta - 3\cos\theta)$ $\Rightarrow 8c^4 + 4c^3 - 8c^2 - 3c + 1 = 0$ (*) | M1 | Shows roots of equation |
| $\Rightarrow \cos\frac{1}{7}\pi,\ \cos\frac{3}{7}\pi,\ \cos\frac{5}{7}\pi,\ -1$ are the roots (AG) | A1 | Part total 2 |
| Sum of roots of (*) are $\frac{-4}{8} = -\frac{1}{2}$ $\Rightarrow$ Result (AG) | M1A1 | States sum of roots; part total 2; **Total [9]** |
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6 Use de Moivre's theorem to show that
$$\cos 4 \theta = 8 \cos ^ { 4 } \theta - 8 \cos ^ { 2 } \theta + 1$$
Without using a calculator, verify that $\cos 4 \theta = - \cos 3 \theta$ for each of the values $\theta = \frac { 1 } { 7 } \pi , \frac { 3 } { 7 } \pi , \frac { 5 } { 7 } \pi , \pi$.
Using the result $\cos 3 \theta = 4 \cos ^ { 3 } \theta - 3 \cos \theta$, show that the roots of the equation
$$8 c ^ { 4 } + 4 c ^ { 3 } - 8 c ^ { 2 } - 3 c + 1 = 0$$
are $\cos \frac { 1 } { 7 } \pi , \cos \frac { 3 } { 7 } \pi , \cos \frac { 5 } { 7 } \pi , - 1$.
Deduce that $\cos \frac { 1 } { 7 } \pi + \cos \frac { 3 } { 7 } \pi + \cos \frac { 5 } { 7 } \pi = \frac { 1 } { 2 }$.
\hfill \mbox{\textit{CAIE FP1 2012 Q6 [9]}}