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


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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Is $m$ an integer? How about writing $\csc=\frac{1}{\sin}$ and expanding the angle sums? Just a thought. – Ross Millikan Jun 20 '11 at 16:38
I modified the sum again. – Venk Jun 20 '11 at 16:41
$\csc(\theta+\pi) = -\csc(\theta)$ so there is some cancellation... – GEdgar Jun 20 '11 at 16:53
@gedgar here it is pi/4 instead of pi. – Venk Jun 20 '11 at 16:59
@GEdgar: so the $m=1$ term is the same as the $m=5$ term, and also for $2$ and $6$ – Ross Millikan Jun 20 '11 at 17:06

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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I don't understand what you are doing. $\sum_{135}$ where is that coming from – user9413 Jun 27 '11 at 11:57
@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. – AlbertH Jun 27 '11 at 12:20

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