Find the value of $x$ such that $$\dfrac{(x+\alpha)^n-(x+\beta)^n}{\alpha-\beta} = \dfrac{\sin n\theta}{\sin ^n \theta}$$ when $\alpha$ and $\beta$ are the roots of $t^2 − 2t + 2 = 0$.
Here is my attempt:
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2Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. – José Carlos Santos Jan 09 '18 at 11:01
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@JoséCarlosSantos I made an attempt https://imgur.com/a/yWVKb – Vivek Jan 09 '18 at 11:14
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@Vivek Always include your effort beforehand, else question may get closed and your effort goes in vain. – jonsno Jan 09 '18 at 11:17
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@samjoe Ok... Sorry, I didn't know this... I am new here – Vivek Jan 09 '18 at 11:18
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Hint:
WLOG let $\alpha=1+i,\beta+1-i\implies\alpha+\beta=?,\alpha-\beta=?$
Using How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i*\sin(\varphi)$?, $$e^{ix}-e^{-ix}=2i\sin x$$
$$\implies\dfrac{\sin(n\theta)}{\sin^n\theta}=\dfrac1{2i}\cdot\left(\left(\dfrac{e^{i\theta}}{\sin\theta}\right)^n-\left(\dfrac{e^{-i\theta}}{\sin\theta}\right)^n\right)$$
Can you take it from here?
lab bhattacharjee
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@samjoe, $$1+i+x=\dfrac{e^{i\theta}}{\sin\theta}=\cot\theta+i\iff x=?$$ – lab bhattacharjee Jan 09 '18 at 11:30
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@Lab I got this only, but not sure how we get any value for this. Is it only $x=\cot(\theta) - 1$ ? I thought some numerical value was answer! – jonsno Jan 09 '18 at 11:32
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WLOG let $\alpha=1+i,\beta+1-i\implies\alpha+\beta=?,\alpha-\beta=?$
Let $2\cot t=x+\alpha+x+\beta=2(x+1)$
$$(x+\alpha)^n-(x+\beta)^n=\dfrac{2i\sin nt}{\sin^nt}=\dfrac{(\alpha-\beta)\sin nt}{\sin^nt}$$
using de Moivre's theorem
lab bhattacharjee
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why have you done this? why are you writing $2\cot t=x+\alpha+x+\beta=2(x+1)$ ? – Vivek Jan 09 '18 at 11:41
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@Vivek, So that $x+\alpha, x+\beta$ becomes of the form $$a\pm ib$$ – lab bhattacharjee Jan 09 '18 at 11:51
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I am not following you.. $x+\alpha=x+1+i, x+\beta=x+1-i$ which is in the form $x+1\pm i$ so what? – Vivek Jan 09 '18 at 12:06
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@Vivek, Then we can employ https://en.wikipedia.org/wiki/De_Moivre%27s_formula after replacing $x+1=\cot t$ – lab bhattacharjee Jan 09 '18 at 12:07
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1@Vivek, $$x+\alpha=\cot t-1+1+i=\dfrac{\cos t+i\sin t}{\sin t}$$ right? $$x+\beta=?$$ – lab bhattacharjee Jan 09 '18 at 12:54
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yes I get it... I wanted to ask, your second solution can't be an independent solution, right? Because you used the answer from first solution that $x=cot(t) -1$ right? – Vivek Jan 10 '18 at 05:05
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@Vivek,IMHO, this is independent & more logical. I've used introduced $\cot t$ to utilize de Moivre's theorem – lab bhattacharjee Jan 10 '18 at 05:18
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@Vivek, $$\cot t$$ is the arithmetic mean hence equi-distant from both – lab bhattacharjee Jan 10 '18 at 05:38