MHB Solving $\sum_{m=1}^{6}\csc \theta$ for $\theta$

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The equation to solve is the sum of cosecants expressed as $$\sum_{m=1}^{6}\csc \left\{ \theta +\frac{(m-1)\pi}{4}\right\}\csc \left\{ \theta +\frac{m\pi}{4}\right\}=4\sqrt{2}$$ for \(0<\theta < \frac{\pi}{2}\). Simplifying this expression leads to the equation \(\sin 2\theta=\frac{1}{2}\). The solutions to this equation are \(\theta=\frac{\pi}{12}\) and \(\theta=\frac{5\pi}{12}\). Numerical methods can also confirm these solutions. The discussion emphasizes the ease of finding and verifying the solutions.
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For $0<\theta < \frac{\pi}{2}$, find the solution(s) of

$$\sum_{m=1}^{6}\csc \left\{ \theta +\frac{(m-1)\pi}{4}\right\}\csc \left\{ \theta +\frac{m\pi}{4}\right\}=4\sqrt{2}$$
 
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sbhatnagar said:
For $0<\theta < \frac{\pi}{2}$, find the solution(s) of

$$\sum_{m=1}^{6}\csc \left\{ \theta +\frac{(m-1)\pi}{4}\right\}\csc \left\{ \theta +\frac{m\pi}{4}\right\}=4\sqrt{2}$$

It is fairly easy to find the solutions numerically and then to verify that they are indeed solutions, IIRC the solutions are \(\pi/12\) and \(5 \pi/12\)

CB
 
sbhatnagar said:
For $0<\theta < \frac{\pi}{2}$, find the solution(s) of

$$\sum_{m=1}^{6}\csc \left\{ \theta +\frac{(m-1)\pi}{4}\right\}\csc \left\{ \theta +\frac{m\pi}{4}\right\}=4\sqrt{2}$$

Hi sbhatnagar, :)

\[\sum_{m=1}^{6}\csc \left\{ \theta +\frac{(m-1)\pi}{4}\right\}\csc \left\{ \theta +\frac{m\pi}{4}\right\}=4\sqrt{2}\]

Expanding the sum and simplification yields,

\[\frac{2(\sin\theta+\cos\theta)}{\sin\left(\theta+ \frac{\pi}{4}\right)}+\frac{\sin\theta-\cos\theta}{\cos\left(\theta+\frac{\pi}{4}\right)}=4\sqrt{2}\sin\theta\cos\theta\]

\[\Rightarrow\sin 2\theta=\frac{1}{2}\]

\[\therefore \theta=\frac{\pi}{12}\mbox{ or }\theta=\frac{5\pi}{12}\]

Kind Regards,
Sudharaka.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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