# Question 8:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(\cos\theta + i\sin\theta)^5 = \cos5\theta + i\sin5\theta$ | B1 | Applies de Moivre correctly. Need not see full statement, but must be correctly applied |
| Use binomial theorem to expand $(\cos\theta + i\sin\theta)^5$: $= \cos^5\theta + 5\cos^4\theta(i\sin\theta) + \frac{5\times4}{2!}\cos^3\theta(i\sin\theta)^2 + \frac{5\times4\times3}{3!}\cos^2\theta(i\sin\theta)^3 + \frac{5\times4\times3\times2}{4!}\cos\theta(i\sin\theta)^4 + (i\sin\theta)^5$ | M1 | Use binomial theorem to expand. May only show imaginary parts - ignore errors in real parts. Binomial coefficients must be evaluated |
| $= \cos^5\theta + 5i\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10i\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + i\sin^5\theta$ | A1 | Simplify coefficients to obtain a simplified result with all imaginary terms correct |
| $\sin5\theta = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta$ | M1 | Equate imaginary parts and obtain an expression for $\sin5\theta$ in terms of powers of $\sin\theta$. No $\cos\theta$ now |
| $= 5(1-\sin^2\theta)^2\sin\theta - 10(1-\sin^2\theta)\sin^3\theta + \sin^5\theta$ | | |
| $= 5(1-2\sin^2\theta+\sin^4\theta)\sin\theta - 10(1-\sin^2\theta)\sin^3\theta + \sin^5\theta$ | | |
| $\sin5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta$ | A1* | Correct **given** result from fully correct working with at least one intermediate line with $(1-\sin^2\theta)^2$ expanded. Must see both sides. A0 if equating of imaginary terms is not clearly implied |

**Alternative method:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Using $z - \frac{1}{z}$: $z^5 - \frac{1}{z^5} = 2i\sin5\theta$ | B1 | |
| Binomial expansion of $\left(z - \frac{1}{z}\right)^5$ | M1 | |
| $32\sin^5\theta = 2\sin5\theta - 10\sin3\theta + 20\sin\theta$ | A1 | |
| Uses double angle formulae to obtain $\sin3\theta = 3\sin\theta - 4\sin^3\theta$ and uses it in expansion | M1 | |
| $\sin5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta$ | A1* | |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $x = \sin\theta$, $16x^5 - 20x^3 + 5x = -\frac{1}{5} \Rightarrow \sin5\theta = \ldots$ | M1A1 | Use substitution $x = \sin\theta$ and attempt to use result from (a) to obtain value for $\sin5\theta$. Note: answers only with no working score no marks |
| $\Rightarrow \theta = \frac{1}{5}\sin^{-1}\left(\pm\frac{1}{5}\right) = 38.306$ (or $-2.307, 69.692, 110.306, 141.693, 182.306$) (or in radians $-0.0402\ldots, 0.6685\ldots, 1.216\ldots, 1.925\ldots, 2.473\ldots$) | dM1 | Proceeds to apply arcsin and divide by 5 to obtain at least one value for $\theta$ |
| Two of (awrt) $x = \sin\theta = -0.963, -0.555, -0.040, 0.620, 0.938$ | A1 | |
| All of (awrt) $x = \sin\theta = -0.963, -0.555, -0.040, 0.620, 0.938$ | A1 | |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_0^{\frac{\pi}{4}}(4\sin^5\theta - 5\sin^3\theta - 6\sin\theta)\,d\theta = \int_0^{\frac{\pi}{4}}\frac{1}{4}(\sin5\theta - 5\sin\theta) - 6\sin\theta\,d\theta$ | M1 | Use previous work to change integrand into a function that can be integrated |
| $= \left[\frac{1}{4}\left(-\frac{1}{5}\cos5\theta + 5\cos\theta\right) + 6\cos\theta\right]_0^{\frac{\pi}{4}} = \left[-\frac{1}{20}\cos5\theta + \frac{29}{4}\cos\theta\right]_0^{\frac{\pi}{4}}$ | A1 | Correct result after integrating. Any limits shown can be ignored |
| $\frac{1}{4}\left[-\frac{1}{5}\cos\frac{5\pi}{4} + 5\cos\frac{\pi}{4} - \left(-\frac{1}{5}+5\right)\right] + 6\cos\frac{\pi}{4} - 6$ | | |
| $= \frac{1}{4}\left[\frac{1}{5}\times\frac{1}{\sqrt{2}} + \frac{5}{\sqrt{2}} - 4\frac{4}{5}\right] + \frac{6}{\sqrt{2}} - 6$ | dM1 | Substitute given limits, subtract and use exact numerical values for trig functions |
| $= \dfrac{73\sqrt{2}}{20} - \dfrac{36}{5}$ | A1 | Final answer correct |