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Show that $${\left(\frac{1+\sin{\frac{\pi}{8}}+i{\cos{\frac{\pi}{8}}}}{1+\sin{\frac{\pi}{8}}-i{\cos{\frac{\pi}{8}}}}\right)}^{\frac{8}{3}}=-1$$

My Try

I know that I have to apply de Moivre's formula to simplify this further. But it can only be used for integers. I tried simplifying this expression but its going no where,

${\left(\frac{\sin^2{\frac{\pi}{16}}+\cos^2{\frac{\pi}{16}} +2\cos{\frac{\pi}{16}}\sin{\frac{\pi}{16}}+i{\cos{\frac{\pi}{8}}}}{\sin^2{\frac{\pi}{16}}+\cos^2{\frac{\pi}{16}} +2\cos{\frac{\pi}{16}}\sin{\frac{\pi}{16}}-i{\cos{\frac{\pi}{8}}}}\right)}^{\frac{8}{3}}$

Can someone please give me a hint to work this out? Thank you!

emil
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2 Answers2

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Hint

Let $\dfrac\pi8=\dfrac\pi2-2t$

$$S=\left(\dfrac{1+\cos2t+i\sin2t}{1+\cos2t-i\sin2t}\right)^{8/3}$$

$$=\left(\dfrac{\cos t+i\sin t}{\cos t-i\sin t}\right)^{8/3}$$ as $\cos t\ne0$

Either utilise Proof for de Moivre's Formula

Or using Intuition behind euler's formula,

$$S=(e^{2it})^{8/3}$$

So, one of the three values of $S$ is

$$e^{16it/3}$$

Can you take it from here?

lab bhattacharjee
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  • Thank you very for your response. I was able to complete the missing steps as follows,$$=\left(\dfrac{\cos t+i\sin t}{\cos t-i\sin t}\right)^{8/3}$$ $$=\left(\dfrac{(\cos t+i\sin t)^2}{\cos^2 t+\sin^2 t}\right)^{8/3}$$ $$=\left(\cos t+i\sin t\right)^{16/3}$$ By Demovier rule:

    $$=\cos \frac{16t}{3}+i\sin\frac{16t}{3}$$ since $\frac{16t}{3}=\pi$ $$=\cos \pi+i\sin\pi$$ $$=-1$$

     – emil Nov 16 '20 at 12:17
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HINT:

The simplest thing I could think of is using the given relation:

$$1+e^{i\alpha} = 2\cos(\frac{\alpha}{2})e^{i\frac{\alpha}{2}}$$

Simple $\frac{\pi}{2}$ addition/subtration in $\sin{x}$ or $\cos{x}$ can bring both the terms in the numerator and denominator into either

$$1+e^{ix}$$ or $$1+e^{-ix}$$

form. From there you can apply the relation I have given above.

Aman Kumar
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