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Use De Moivre's theorem to show that $$\cos4x=8\sin^4x-8\sin^2x+1$$ Hence show the one of the roots of the equation $8z^4-8z^2+1=0$ is $\sin\frac{\pi}{8}$ and express the other roots in polar form. Deduce that $\sin \frac{\pi}{8}=\frac{1}{2}\sqrt{2-\sqrt{2}}$ and find an exact expression for $\sin \frac{11}{8}\pi$.

My attempt,

$\cos4x=Re(\cos4x+i\sin4x)$

$=Re(\cos x+i\sin x)^4$

$=Re(c^4+4c^3is+6c^2i^2s^2+4ci^3s^3+i^4s^4)$

$=c^4-6c^2s^2+s^4$

So, $\cos4x=\cos^4x-6\cos^2x\sin^2x+\sin^4x$

$=(1-\sin^2x)^2-6(1-\sin^2x)\sin^2x+sin^4x$

$1-2\sin^2x+\sin^4x-6\sin^2x+6\sin^4x+\sin^4x$

$=8\sin^4x-8\sin^2x+1$

How to proceed then?

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  • $\begingroup$ Make use of the relation $\sin^2x+\cos^2x=1$. $\endgroup$ – Kwin van der Veen Jun 9 '15 at 10:26
  • $\begingroup$ How to proceed then? $\endgroup$ – MathLOL Jun 9 '15 at 10:28
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let $cosx =c $, $ sinx = s$

$cos4x = (1-s^2)^2-6(1-ss)\cdot s^2+s^4$

$= 1 -2s^2 +s^4 - 6s^2 +6s^4 +s^4$

$= 8s^4 -8s^2 +1$

$cos 4x = 0 = 8s^4 -8s^2 +1$

$cos \frac{\pi}{2} =0$

$x= \frac{\pi}{8}$

so $sin \frac{\pi}{8}$ is a solution

Look at the roots of $8s^4-8s^2 +1$ and use the quadratic equation to solve we see that there is only one solution in range $(0,0.5)$ which is $\frac{\sqrt{(2-\sqrt{(2)})}}{2}$ so this has to be sin $\frac{\pi}{8}$

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  • $\begingroup$ How to find $\sin \frac{11}{8}$ then? $\endgroup$ – MathLOL Jun 9 '15 at 10:44
  • $\begingroup$ cos( 4 *11/8 pi ) =0 so look for the other solution of 0 = 8s^4 -8s^2 +1 in the range (0,1) $\endgroup$ – RowanS Jun 9 '15 at 10:47
  • $\begingroup$ How you got $\cos (4\cdot \frac{11}{8}\pi)$ ? $\endgroup$ – MathLOL Jun 9 '15 at 10:50
  • $\begingroup$ cos 4x = 0 = 8s^4 -8s^2 +1 $\endgroup$ – RowanS Jun 9 '15 at 10:52
  • $\begingroup$ @MathLOL $\cos\left(\pi(1/2+n)\right)=0$ for any integer $n$. $\endgroup$ – Kwin van der Veen Jun 9 '15 at 13:49

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