Question of the day 1.09

AA HL · Complex Numbers · Trigonometry

(a) Use de Moivre’s theorem to prove that

$\cos 5\theta = \cos\theta \left( 16\cos^4\theta-20\cos^2\theta+5 \right).$

(b) By considering the equation $\cos 5\theta=0$ , show that

$\cos^2\left(\frac{\pi}{10}\right) = \frac{5+\sqrt5}{8}.$

Hint

For part (a), consider the real part of

$(\cos\theta+i\sin\theta)^5.$

For part (b), notice that

$16\cos^4\theta-20\cos^2\theta+5$

is a quadratic in $u=\cos^2\theta$ .

Solution

(a)

By de Moivre’s theorem,

$\cos 5\theta+i\sin 5\theta = (\cos\theta+i\sin\theta)^5.$

Taking real parts gives

$\begin{aligned} \cos 5\theta &= \operatorname{Re}(\cos\theta+i\sin\theta)^5\\ &= \cos^5\theta -10\cos^3\theta\sin^2\theta +5\cos\theta\sin^4\theta. \end{aligned}$

Using $\sin^2\theta=1-\cos^2\theta$ ,

$\begin{aligned} \cos 5\theta &= \cos^5\theta -10\cos^3\theta(1-\cos^2\theta) +5\cos\theta(1-\cos^2\theta)^2\\ &= \cos^5\theta -10\cos^3\theta +10\cos^5\theta\\ &\qquad +5\cos\theta -10\cos^3\theta +5\cos^5\theta\\ &= 16\cos^5\theta -20\cos^3\theta +5\cos\theta\\ &= \cos\theta \left( 16\cos^4\theta -20\cos^2\theta +5 \right). \end{aligned}$

Hence

$\boxed{ \cos 5\theta = \cos\theta \left( 16\cos^4\theta-20\cos^2\theta+5 \right) }.$

(b)

Consider

$\cos 5\theta=0.$

Five solutions in the interval $0<\theta<\pi$ are

$\theta= \frac{\pi}{10}, \frac{3\pi}{10}, \frac{5\pi}{10}, \frac{7\pi}{10}, \frac{9\pi}{10}.$

Using the identity from part (a),

$\cos\theta \left( 16\cos^4\theta-20\cos^2\theta+5 \right)=0.$

Therefore either

$\cos\theta=0$

$16\cos^4\theta-20\cos^2\theta+5=0.$

The first equation gives

$\theta=\frac{\pi}{2} = \frac{5\pi}{10}.$

For the remaining four solutions, let

$u=\cos^2\theta.$

Then

$16u^2-20u+5=0.$

Using the quadratic formula,

$\begin{aligned} u &= \frac{20\pm\sqrt{400-320}}{32}\\ &= \frac{20\pm\sqrt{80}}{32}\\ &= \frac{20\pm4\sqrt5}{32}\\ &= \frac{5\pm\sqrt5}{8}. \end{aligned}$