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For $ 0 <\theta<\frac{\pi}{2}$ find the solution of

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

I thought of solving this as the angles form an A.P , But the given sum does not come under any standard type such as the sum of the sines or cosines of the angles in an A.P.So I am unable to proceed further.

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    @GEdgar: so the $m=1$ term is the same as the $m=5$ term, and also for $2$ and $6$2011-06-20

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The left side of the equation can be rewritten as: $ \Delta = \sum_{1 3 5} \csc\left(\theta+\frac{m\pi}{4}\right) \cdot \left( \csc\left(\theta+\frac{(m-1)\pi}{4}\right) + \csc\left(\theta+\frac{(m+1)\pi}{4}\right) \right) $ Now using the formulas $ \sin u +\sin v = 2\sin\left(\frac {u + v} 2\right) \cdot \cos \left(\frac {u - v} 2\right)$ and $ \sin u \sin v = \frac 1 2 \left(\cos (u - v) - \cos (u + v)\right)$ we have $ \begin{aligned}\csc\left(\theta+\frac{(m-1)\pi}{4}\right) + \csc\left(\theta+\frac{(m+1)\pi}{4}\right) &= \frac {\sin\left(\theta+\frac{(m-1)\pi}{4}\right) + \sin\left(\theta+\frac{(m+1)\pi}{4}\right)} {\sin\left(\theta+\frac{(m-1)\pi}{4}\right) \cdot \sin\left(\theta+\frac{(m+1)\pi}{4}\right)}\\ &= \frac {4 \sin\left(\theta + \frac {m\pi} {4}\right) cos\left( \frac \pi 4\right)}{\cos\left(\frac \pi 2 \right) - \cos\left(2\theta + \frac {m\pi} {2} \right)}\end{aligned}$ So $\Delta$ becomes $\Delta = -2\sqrt 2\sum_{1 3 5} \frac 1 {\cos\left(2\theta + \frac {m\pi} {2} \right)} = -2\sqrt 2 \left( -\frac 1 {\sin 2\theta} +\frac 1 {\sin 2\theta} -\frac 1 {\sin 2\theta}\right) = 2\sqrt 2 \frac 1 {\sin 2\theta}$ I hope there are no typos... ;)

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    @Chandru: With $\sum_{1 3 5}$ I indicate the summation with index $m$ varying in the set $\{1 3 5\}$. "My" summand with $m=1$ equals the sum of the first two original summands and so on.2011-06-27