10. (a) Given that \(z = e ^ { \mathrm { i } \theta }\), show that
$$z ^ { n } - \frac { 1 } { z ^ { n } } = 2 \mathrm { i } \sin n \theta$$
where \(n\) is a positive integer.
Show mark scheme
Show mark scheme source
Question 10:
Part (a)
Answer Marks
Guidance
Answer Marks
Guidance
\(z^n = e^{in\theta} = (\cos n\theta + i\sin n\theta)\), \(z^{-n} = e^{-in\theta} = (\cos n\theta - i\sin n\theta)\) M1
Completion: \(z^n - \overline{z^n} = 2i\sin n\theta\) (*) AG A1
2 marks total
Part (b)
Answer Marks
Guidance
Answer Marks
Guidance
\(\left(z - \dfrac{1}{z}\right)^5 = z^5 - 5z^3 + 10z - \dfrac{10}{z} + \dfrac{5}{z^3} - \dfrac{1}{z^5}\) M1A1
\(= \left(z^5 - \dfrac{1}{z^5}\right) - 5\left(z^3 - \dfrac{1}{z^3}\right) + 10\left(z - \dfrac{1}{z}\right)\) \((2i\sin\theta)^5 = 32i\sin^5\theta = 2i\sin 5\theta - 10i\sin 3\theta + 20i\sin\theta\) M1A1
\(\Rightarrow \sin^5\theta = \dfrac{1}{16}(\sin 5\theta - 5\sin 3\theta + 10\sin\theta)\) (*) AG A1
5 marks total
Part (c)
Answer Marks
Guidance
Answer Marks
Guidance
Finding \(\sin^5\theta = \tfrac{1}{4}\sin\theta\) M1
\(\theta = 0,\ \pi\) (both) B1
\((\sin^4\theta = \tfrac{1}{4}) \Rightarrow \sin\theta = \pm\dfrac{1}{\sqrt{2}}\) M1
\(\theta = \dfrac{\pi}{4},\ \dfrac{3\pi}{4};\ \dfrac{5\pi}{4},\ \dfrac{7\pi}{4}\) A1;A1
5 marks total [12]
Copy
# Question 10:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z^n = e^{in\theta} = (\cos n\theta + i\sin n\theta)$, $z^{-n} = e^{-in\theta} = (\cos n\theta - i\sin n\theta)$ | M1 | |
| Completion: $z^n - \overline{z^n} = 2i\sin n\theta$ (*) AG | A1 | 2 marks total |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(z - \dfrac{1}{z}\right)^5 = z^5 - 5z^3 + 10z - \dfrac{10}{z} + \dfrac{5}{z^3} - \dfrac{1}{z^5}$ | M1A1 | |
| $= \left(z^5 - \dfrac{1}{z^5}\right) - 5\left(z^3 - \dfrac{1}{z^3}\right) + 10\left(z - \dfrac{1}{z}\right)$ | | |
| $(2i\sin\theta)^5 = 32i\sin^5\theta = 2i\sin 5\theta - 10i\sin 3\theta + 20i\sin\theta$ | M1A1 | |
| $\Rightarrow \sin^5\theta = \dfrac{1}{16}(\sin 5\theta - 5\sin 3\theta + 10\sin\theta)$ (*) AG | A1 | 5 marks total |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Finding $\sin^5\theta = \tfrac{1}{4}\sin\theta$ | M1 | |
| $\theta = 0,\ \pi$ (both) | B1 | |
| $(\sin^4\theta = \tfrac{1}{4}) \Rightarrow \sin\theta = \pm\dfrac{1}{\sqrt{2}}$ | M1 | |
| $\theta = \dfrac{\pi}{4},\ \dfrac{3\pi}{4};\ \dfrac{5\pi}{4},\ \dfrac{7\pi}{4}$ | A1;A1 | 5 marks total **[12]** |
---
Show LaTeX source
Copy
10. (a) Given that $z = e ^ { \mathrm { i } \theta }$, show that
$$z ^ { n } - \frac { 1 } { z ^ { n } } = 2 \mathrm { i } \sin n \theta$$
where $n$ is a positive integer.\\
(b) Show that
$$\sin ^ { 5 } \theta = \frac { 1 } { 16 } ( \sin 5 \theta - 5 \sin 3 \theta + 10 \sin \theta )$$
(c) Hence solve, in the interval $0 \leq \theta < 2 \pi$,
$$\sin 5 \theta - 5 \sin 3 \theta + 6 \sin \theta = 0$$
(5)(Total 12 marks)\\
\hfill \mbox{\textit{Edexcel FP2 2005 Q10 [12]}}