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Let $x_s = \sin(\theta+\frac{2\pi s}{3})$ and $y_s = 1+\cos(\frac{2\pi s}{3})$, $s=0,1,2$.

Define $f(\theta) = \sum_{s=0}^2 x_s\ln y_s$.

Is there any method to derive roots of $f(\theta)$. I have run a simulation on it, and found that $\theta=0$ is a solution. But I am unable to see how to analytically obtain it.

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This would be better suited to math.stackexchange.com –  Did Jun 19 '11 at 8:52

1 Answer 1

up vote 4 down vote accepted

By trigonometry, $$f(\theta)= \log (2)\left( \sin\theta+\cos(\pi/6-\theta)-\cos(\pi/6+\theta) \right)=\log(4)\sin\theta.$$


For the revised problem, we have this: $f(\theta)$ is $2\pi/3$-periodic, and $f(\theta)$ is odd, so it suffices to find the roots between $0$ and $\pi/3$ (both of which are themselves roots). Plotting indicates that $f(\theta)$ is unimodal on this interval, $f'(\theta)$ is strictly increasing, and $f''(\theta)$ is strictly increasing, and $f'''(\theta)\geq 6$. Each of these observations follows from the one after it (sometimes needing to also evaluate at $\theta=0$), and the last one seems easiest to prove. Not elegant, certainly, but it should get the job done.

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Thank you. But I didn't express my question correctly. The $\theta$ is in $y_s$ too. Anyway, thanks for your reply. –  Victor Jun 21 '11 at 4:08
    
In $y_s$ where? –  Kevin O'Bryant Jun 21 '11 at 11:42
    
It should be $y_s=1+\cos(\theta+\frac{2\pi s}{3})$ –  Victor Jun 30 '11 at 16:44

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