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.
 
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