2022TrialShort answer

Q16 (a) (ii)

2 marks

Question

Hence show that
$\cos^7\theta=\frac1{64}\left(\cos7\theta+7\cos5\theta+21\cos3\theta+35\cos\theta\right).$

Try the question before revealing the answer.

Solution

(ii)
$\left(z+z^{-1}\right)^7=z^7+7z^5+21z^3+35z+\frac{35}{z}+\frac{21}{z^3}+\frac{7}{z^5}+\frac{1}{z^7}$

$\left(z+\frac{1}{z}\right)^7=\left(z^7+\frac{1}{z^7}\right)+7\left(z^5+\frac{1}{z^5}\right)+21\left(z^3+\frac{1}{z^3}\right)+35\left(z+\frac{1}{z}\right)$

$\therefore (2\cos\theta)^7=2\cos7\theta+14\cos5\theta+42\cos3\theta+70\cos\theta$
$128\cos^7\theta=2\cos7\theta+14\cos5\theta+42\cos3\theta+70\cos\theta$
$\cos^7\theta=\frac{1}{64}\left[\cos7\theta+7\cos5\theta+21\cos3\theta+35\cos\theta\right]$

as required.

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Complex Numbers

Other representations of complex numbersUse Euler's formula and convert between Cartesian, polar and exponential forms to solve problems.

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