7.(a)Use de Moivre's theorem to show that
$$\cos 7 \theta \equiv 64 \cos ^ { 7 } \theta - 112 \cos ^ { 5 } \theta + 56 \cos ^ { 3 } \theta - 7 \cos \theta$$
(b)Hence find the four distinct roots of the equation
$$64 x ^ { 7 } - 112 x ^ { 5 } + 56 x ^ { 3 } - 7 x + 1 = 0$$
giving your answers to 3 decimal places where necessary.
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Question 7:
Part (a):
Answer Marks
Guidance
Answer/Working Mark
Guidance
\((\cos\theta + i\sin\theta)^7 = \cos^7\theta + \binom{7}{1}\cos^6\theta\, i\sin\theta + \binom{7}{2}\cos^5\theta(i\sin\theta)^2 + \ldots\) M1
Attempts to expand \((\cos\theta + i\sin\theta)^7\) including a recognisable attempt at binomial coefficients (may only see real terms)
\((\cos 7\theta =)\ c^7 + {}^7C_2c^5i^2s^2 + {}^7C_4c^3i^4s^4 + {}^7C_6ci^6s^6\) M1
Identifies real terms with \(\cos 7\theta\)
\(= c^7 - 21c^5s^2 + 35c^3s^4 - 7cs^6\) A1
Correct expression with coefficients evaluated and \(i\)'s dealt with correctly
\(= c^7 - 21c^5(1-c^2) + 35c^3(1-c^2)^2 - 7c(1-c^2)^3\) M1
Replaces \(\sin^2\theta\) with \(1 - \cos^2\theta\) used anywhere in expansion
\(= 22c^7 - 21c^5 + 35c^3(1 - 2c^2 + c^4) - 7c(1 - 3c^2 + 3c^4 - c^6)\) M1
Applies expansions of \((1-\cos^2\theta)^2\) and \((1-\cos^2\theta)^3\) to their expression
\(= 64\cos^7\theta - 112\cos^5\theta + 56\cos^3\theta - 7\cos\theta\) A1
Correct expression obtained with no errors
Alternative 1:
Answer Marks
Guidance
Answer/Working Mark
Guidance
\(\left(z + \frac{1}{z}\right)^7 = z^7 + \binom{7}{1}z^6\frac{1}{z} + \binom{7}{2}z^5\frac{1}{z^2} + \ldots\) M1
Attempts to expand \(\left(z+\frac{1}{z}\right)^7\) including binomial coefficients
\((2\cos\theta)^7 = 2\cos 7\theta + 7(2\cos 5\theta) + 21(2\cos 3\theta) + 35(2\cos\theta)\) M1A1
M1: Uses \(z^n + \frac{1}{z^n} = 2\cos n\theta\) at least once (including \(n=1\)); A1: Correct expression in terms of cos
\(128\cos^7\theta = 2\cos 7\theta + 14(16\cos^5\theta - 20\cos^3\theta + 5\cos\theta) + 42(4\cos^3\theta - 3\cos\theta) + 70\cos\theta\) M1M1
M1: Correct method to find \(\cos 5\theta\) in terms of \(\cos\theta\); M1: Correct method to find \(\cos 3\theta\) in terms of \(\cos\theta\)
\(\cos 7\theta = 64\cos^7\theta - 112\cos^5\theta + 56\cos^3\theta - 7\cos\theta\) A1
Part (b):
Answer Marks
Guidance
Answer/Working Mark
Guidance
\(\cos 7\theta + 1 = 0 \Rightarrow \cos 7\theta = -1\) B1
\(\cos 7\theta = -1\) (\(\cos 7x = -1\) is B0)
\(7\theta = \pm 180°, \pm 540°, \pm 900°, \pm 1260°, \ldots\) or \(7\theta = \pm\pi, \pm 3\pi, \pm 5\pi, \pm 7\pi, \ldots\) M1
At least one correct value for \(7\theta\); condone use of \(7x\) here
\(\theta = \pm\frac{180}{7}, \pm\frac{540}{7}, \pm\frac{900}{7}, \pm\frac{1260}{7},\ldots\) or \(\theta = \pm\frac{\pi}{7}, \pm\frac{3\pi}{7}, \pm\frac{5\pi}{7}, \pm\frac{7\pi}{7},\ldots\) M1
Divides by 7 and attempts at least one value for \(\cos\theta\); condone use of \(x\) for \(\theta\)
\(x = \cos\theta = 0.901, 0.223, -1, -0.623\) A1A1
A1: Awrt 2 correct values for \(x\); A1: Awrt all 4 \(x\) values correct and no extras
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# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(\cos\theta + i\sin\theta)^7 = \cos^7\theta + \binom{7}{1}\cos^6\theta\, i\sin\theta + \binom{7}{2}\cos^5\theta(i\sin\theta)^2 + \ldots$ | M1 | Attempts to expand $(\cos\theta + i\sin\theta)^7$ including a recognisable attempt at binomial coefficients (may only see real terms) |
| $(\cos 7\theta =)\ c^7 + {}^7C_2c^5i^2s^2 + {}^7C_4c^3i^4s^4 + {}^7C_6ci^6s^6$ | M1 | Identifies real terms with $\cos 7\theta$ |
| $= c^7 - 21c^5s^2 + 35c^3s^4 - 7cs^6$ | A1 | Correct expression with coefficients evaluated and $i$'s dealt with correctly |
| $= c^7 - 21c^5(1-c^2) + 35c^3(1-c^2)^2 - 7c(1-c^2)^3$ | M1 | Replaces $\sin^2\theta$ with $1 - \cos^2\theta$ used anywhere in expansion |
| $= 22c^7 - 21c^5 + 35c^3(1 - 2c^2 + c^4) - 7c(1 - 3c^2 + 3c^4 - c^6)$ | M1 | Applies expansions of $(1-\cos^2\theta)^2$ and $(1-\cos^2\theta)^3$ to their expression |
| $= 64\cos^7\theta - 112\cos^5\theta + 56\cos^3\theta - 7\cos\theta$ | A1 | Correct expression obtained with no errors |
**Alternative 1:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(z + \frac{1}{z}\right)^7 = z^7 + \binom{7}{1}z^6\frac{1}{z} + \binom{7}{2}z^5\frac{1}{z^2} + \ldots$ | M1 | Attempts to expand $\left(z+\frac{1}{z}\right)^7$ including binomial coefficients |
| $(2\cos\theta)^7 = 2\cos 7\theta + 7(2\cos 5\theta) + 21(2\cos 3\theta) + 35(2\cos\theta)$ | M1A1 | M1: Uses $z^n + \frac{1}{z^n} = 2\cos n\theta$ at least once (including $n=1$); A1: Correct expression in terms of cos |
| $128\cos^7\theta = 2\cos 7\theta + 14(16\cos^5\theta - 20\cos^3\theta + 5\cos\theta) + 42(4\cos^3\theta - 3\cos\theta) + 70\cos\theta$ | M1M1 | M1: Correct method to find $\cos 5\theta$ in terms of $\cos\theta$; M1: Correct method to find $\cos 3\theta$ in terms of $\cos\theta$ |
| $\cos 7\theta = 64\cos^7\theta - 112\cos^5\theta + 56\cos^3\theta - 7\cos\theta$ | A1 | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos 7\theta + 1 = 0 \Rightarrow \cos 7\theta = -1$ | B1 | $\cos 7\theta = -1$ ($\cos 7x = -1$ is B0) |
| $7\theta = \pm 180°, \pm 540°, \pm 900°, \pm 1260°, \ldots$ or $7\theta = \pm\pi, \pm 3\pi, \pm 5\pi, \pm 7\pi, \ldots$ | M1 | At least one correct value for $7\theta$; condone use of $7x$ here |
| $\theta = \pm\frac{180}{7}, \pm\frac{540}{7}, \pm\frac{900}{7}, \pm\frac{1260}{7},\ldots$ or $\theta = \pm\frac{\pi}{7}, \pm\frac{3\pi}{7}, \pm\frac{5\pi}{7}, \pm\frac{7\pi}{7},\ldots$ | M1 | Divides by 7 and attempts at least one value for $\cos\theta$; condone use of $x$ for $\theta$ |
| $x = \cos\theta = 0.901, 0.223, -1, -0.623$ | A1A1 | A1: Awrt 2 correct values for $x$; A1: Awrt all 4 $x$ values correct and no extras |
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7.(a)Use de Moivre's theorem to show that
$$\cos 7 \theta \equiv 64 \cos ^ { 7 } \theta - 112 \cos ^ { 5 } \theta + 56 \cos ^ { 3 } \theta - 7 \cos \theta$$
(b)Hence find the four distinct roots of the equation
$$64 x ^ { 7 } - 112 x ^ { 5 } + 56 x ^ { 3 } - 7 x + 1 = 0$$
giving your answers to 3 decimal places where necessary.\\
\hfill \mbox{\textit{Edexcel F2 2018 Q7 [11]}}