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Given thatenter image description here

I can do part of this question so here goes:

Substituting $\theta=\frac{1\pi}{11}$ into LHS of given expression gives $$\cos\frac{1\pi}{11} + \cos\frac{2\pi}{11} + \cos\frac{3\pi}{11} +\ldots+\cos\frac{10\pi}{11} = RHS =\frac{\sin(\frac{21\pi}{22})}{2{\sin(\frac{1\pi}{22})}} - \frac{1}{2}$$ and since $$\sin(\frac{21\pi}{22}) = \sin(\pi-(\frac{21\pi}{22})) = \sin(\frac{1\pi}{22})$$ this makes the RHS equal to $\frac{1}2-\frac{1}{2} = 0$ as required so that $\theta=\frac{1\pi}{11}$ is indeed a root as mentioned.

But I am really stuck on trying to find another root in the interval they mention. Please help.

Thank you kindly.

kimtahe6
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BLAZE
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1 Answers1

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If we rearrange a bit, we get

$$\sin\frac{21\theta}{2} - \sin\frac{\theta}{2}=0$$

Now, use the identity $\sin{x} - \sin{y} = 2\sin\frac{x-y}{2}\cos\frac{x+y}{2}$ and we get

$$2\sin5\theta\cos\frac{11\theta}{2}=0$$

We see that $\theta = \frac{\pi}{5}$ and $\theta = \frac{\pi}{11}$ are both solutions.

Yiyuan Lee
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