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Use de Moivre’s theorem to show that $$\ \cos 6θ = 32\cos^6θ − 48\cos^4θ + 18\cos^2θ − 1 $$. Hence solve the equation $$\ 64x^6 − 96x^4 + 36x^2 − 1 = 0$$ giving each root in the form$\ \cos kπ$.

Attempt

I have completed almost the whole question and obtained an equation $$\ \cos\theta = \cos\frac{k\pi}{9} = -\frac{1}{2}$$ The problem is the equation requires six roots. Whereas the above equation is satisfied by $$\ k= \pm1,\pm2,\pm4,\pm5,\pm7,\pm8$$

My question is

  1. Which six roots I am supposed to choose? (I am aware that both plus-minus values will give the same value for x, but it is also possible for the same value of x to be the root of the equation twice. Like $\ x=1$ is a root of the equation $\ x^2-2x+1$ twice.)

  2. Why is such a complication arising?

mathnoob123
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1 Answers1

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Assuming you have proved $$ \cos 6\theta = 32\cos^6\theta − 48\cos^4\theta + 18\cos^2\theta − 1 $$ you get $$ 32\cos^6\theta − 48\cos^4\theta + 18\cos^2\theta=1+\cos6\theta $$ Now set $x=\cos\theta$, so your equation becomes $$ 2(1+\cos6\theta)-1=0 $$ that is $$ \cos6\theta=-\frac{1}{2} $$ which has the solutions $$ 6\theta=\frac{2\pi}{3}+2k\pi \qquad\text{or}\qquad 6\theta=-\frac{2\pi}{3}+2k\pi $$ You can choose $0\le k\le 5$, because every other solution will repeat values for $x$. On the other hand, you can discard the second set of solutions, for the same reason, because $\cos(-\alpha)=\cos\alpha$.

So you get $$ \frac{\pi}{9},\quad \frac{\pi}{9}+\frac{\pi}{3},\quad \frac{\pi}{9}+\frac{2\pi}{3},\quad \frac{\pi}{9}+\frac{3\pi}{3},\quad \frac{\pi}{9}+\frac{4\pi}{3},\quad \frac{\pi}{9}+\frac{5\pi}{3} $$ and the solutions for $x$ are $$ \cos\frac{\pi}{9},\quad \cos\frac{4\pi}{9},\quad \cos\frac{7\pi}{9},\quad \cos\frac{10\pi}{9},\quad \cos\frac{13\pi}{9},\quad \cos\frac{16\pi}{9} $$ which are pairwise distinct.

To summarize: the equation $\cos6\theta=-1/2$ has twelve solutions in the interval $[0,2\pi)$, but your equation in $x$ has only six, because the solutions in $\theta$ can be grouped in pairs that yield the same value for $\cos\theta$.

egreg
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  • The answer given in my book is k =1,2,4,5,7,8. Can you include how to get these k values in your answer? – mathnoob123 Apr 14 '17 at 21:35
  • And you didn't answer my second question, which was what if the polynomial had two same values of x? In that case, I would have to consider pairs of theta which give the same value of cos theta. So how do I disregard that possibility? – mathnoob123 Apr 14 '17 at 21:37
  • @FaiqRaees The polynomial clearly has distinct roots, because we found six of them. Why your book lists those values of $k$ I don't know. – egreg Apr 14 '17 at 21:45
  • Can you please tell me how did you reach your conclusion that the polynomial has distinct roots? If it were a polynomial that could be reduced to a quadratic, it was easy to work it out using the discriminant rule, however, I am not aware of any techniques that could be used at a cubic equation (which can be reached by using $\ y=x^2$ – mathnoob123 Apr 14 '17 at 21:49
  • @FaiqRaees I don't understand: a degree $6$ polynomial has at most six real roots (counted with multiplicities); we found six distinct roots. That's all. – egreg Apr 14 '17 at 21:50