# Trigonometry - Cosec summation

1. Oct 19, 2013

### Saitama

1. The problem statement, all variables and given/known data
If
$$\csc\frac{\pi}{32}+\csc\frac{\pi}{16}+\csc\frac{\pi}{8}+\csc\frac{\pi}{4}+\csc\frac{\pi}{2}$$
has the value equal to $\cot\frac{\pi}{A}$ then find A.
A)61
B)62
C)63
D)64

2. Relevant equations

3. The attempt at a solution
Writing cosec in terms of sin and taking the LCM to make a common denominator doesn't seem to be of any help.

I can find the value of each term but that would be tedious and of no use.

I honestly cannot figure out how should I proceed here.

Any help is appreciated. Thanks!

Last edited: Oct 19, 2013
2. Oct 19, 2013

### Dick

Try showing cot(x)-cot(2x)=csc(2x).

3. Oct 19, 2013

### Simon Bridge

$\csc(\frac{\pi}{2})$ appears twice.
All the angles are successive doublings of $\pi/32$

4. Oct 19, 2013

### Saitama

Wow! Thanks a lot Dick!

How did you come up with that?

Very sorry for the typo, its $\pi/8$ instead of the second $\pi/2$.

5. Oct 19, 2013

### Dick

I guessed the series must telescope somehow. So somehow cot(2x) must be related to cot(x) with the difference related to a csc. Seems obvious in retrospect, yes?

Last edited: Oct 19, 2013
6. Oct 19, 2013

### Saitama

Yes. I liked the way you came up with cot(x)-cot(2x) and solved the problem in few seconds where I was stuck for a week.

Thank you again! :)

7. Oct 19, 2013

### Dick

You're welcome, but it took me more than a "few seconds". Still keeping that strategy in mind might help in the future. If you've got the sum of a bunch of csc's equaling a cot, then if you can express each csc as a difference of two cot's you might be able to sum the series easily. Substitute any other functions you want for 'csc' and 'cot'.