FP2 2016 8 Edexcel

8

(a)

Set z = e^{i \theta}, then:

\displaystyle z^n + \frac 1 {z^n} = 2 \cos n \theta

for any n \in \mathbb N.

so that:

\displaystyle \left(z + \frac 1 z\right)^5 = 2^5 \cos^5 \theta = 32 \cos^5 \theta

Expanding the LHS using the binomial theorem:

\begin{align*}\left(z + \frac 1 z\right)^5 & = z^5 + \binom 5 1 z^4 z^{-1} + \binom 5 2 z^3 z^{-2} + \binom 5 3 z^2 z^{-3} + \binom 5 4 z z^{-4} + \binom 5 5 z^{-5} \\ & = z^5 + 5z^3 + 10z + \frac {10} z + \frac 5 {z^3} + \frac 1 {z^5} \\ & = \left(z^5 + \frac 1 {z^5}\right) + 5 \left(z^3 + \frac 1 {z^3}\right) + 10 \left(z + \frac 1 z\right) \\ & = 2 \cos 5\theta + 10 \cos 3\theta + 20 \cos \theta\end{align*}

so:

\begin{align*}\cos^5 \theta & \equiv \frac 1 {32} \left(2 \cos 5\theta + 10 \cos 3\theta + 20 \cos \theta\right) \\ & = \frac 1 {16} \cos 5\theta + \frac 5 {16} \cos 3\theta + \frac 5 8 \cos \theta\end{align*}

in particular p = \dfrac 1 {16}, q = \dfrac 5 {16} and r = \dfrac 5 8.

(b)

We have:

\begin{align*}\int_{\frac \pi 6}^{\frac \pi 3} \cos^5 \mathrm d\theta & = \int_{\frac \pi 6}^{\frac \pi 3} \left(\frac 1 {16} \cos 5\theta + \frac 5 {16} \cos 3\theta + \frac 5 8 \cos \theta\right) \mathrm d\theta \\ & = \left[\frac 1 {16 \times 5} \sin 5 \theta + \frac 5 {16 \times 3} \sin 3 \theta + \frac 5 8 \sin \theta\right]_{\frac \pi 6}^{\frac \pi 3} \\ & = \left[\frac 1 {80} \sin 5 \theta + \frac 5 {48} \sin 3 \theta + \frac 5 8\sin \theta\right]_{\frac \pi 6}^{\frac \pi 3} \\ & = \left(\frac 1 {80} \sin \frac {5 \pi} 3 + \frac 5 {48} \sin \frac {3 \pi} 3 + \frac 5 8 \sin \frac \pi 3\right) - \left(\frac 1 {80} \sin \frac {5 \pi} 6 + \frac 5 {48} \sin \frac {3 \pi} 6 + \frac 5 8\sin \frac \pi 6\right) \\ &= \frac {49} {160} \sqrt 3 - \frac {203} {480}\end{align*}

using the common integral:

\displaystyle \int \cos a \theta \mathrm d\theta = \frac 1 a \sin a \theta

for a \ne 0.

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