# Computing $\prod_{x=1}^{44}\left(1-\frac{1}{\tan(x°)}\right)$

I would like to calculate:

$$\prod_{x=1}^{44}\left(1-\frac{1}{\tan(x°)}\right)$$

Here is what I found:

$$1-\frac{1}{\tan(x)}=\frac{\sqrt{2}\sin(x-45°)}{\sin(x)}$$

$$\prod_{x=1}^{44}\left(1-\frac{1}{\tan(x°)}\right)=\prod_{x=1}^{44} \frac{\sqrt{2}\sin(x-45°)}{\sin(x)}=2^{22} \frac{\prod_{x=1}^{44} \sin(x-45°)}{\prod_{x=1}^{44} \sin(x)}$$

$$x\rightarrow 45°-x$$

$$\prod_{x=1}^{44}\left(1-\frac{1}{\tan(x°)}\right)=\frac{\prod_{x=1}^{44} -\sin(x)}{\prod_{x=1}^{44} \sin(x)}=2^{22}$$

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So what is the question? You seem to have your answer. –  Henning Makholm Dec 29 '11 at 22:39
That looks right to me, and my computer confirms it. Nice one, that's quite an elegant result! –  Lopsy Dec 29 '11 at 22:40
This is a variant of math.stackexchange.com/questions/75825/… (perhaps even the same..). –  Srivatsan Dec 29 '11 at 22:45

I don't quite understand what you mean by $x \to 45^\circ-x$, but what you wish to convey is rather intuitive. You'd like to say, that the terms will cancel away, which is easy to see as numerator and denominator are the sines of same arguments but written reverse. So, I hope this helps

EDIT: It's true that this is the age old of trick called Change of variable, but, I don't see a notation that incorporates the change in limit that occurs with the change in variable. This is just a matter of notation and people's way of doing math.

Your Idea is right and you are through.

An alternative approach will be to observe the following:

I'll assume that you know that $$\tan(x+y)=\dfrac{\tan x + \tan y}{1-\tan x \tan y}$$ and point you to the following rather nice way of doing it:

$$\cot(x+y)=\dfrac{1}{\tan(x+y)}=\dfrac{1-\tan x \tan y}{\tan x + \tan y}=\dfrac{\cot x \cot y-1}{\cot x + \cot y}$$

Now, if $x+y = 45^\circ$, then $$\cot x \cot y-1=\cot x + \cot y$$ Tweak this a little to observe, $$(1-\cot x)(1-\cot y)=2$$

Now, see that your product coincides with $22$ such pairs $(1,44); (2, 43);....(22,23)$ and this gives you the answer.

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Exactly, from $\prod_{x=1}^{44} \frac{\sqrt{2}\sin(x-45°)}{\sin(x)}$ you have the answer as $(\sqrt 2)^{44}$. –  Gigili Dec 29 '11 at 22:45
The arguments in the numerator are symmetric to the denominator's, but since the $\sin x$ function is odd, the product of 44 negative terms in the numerator is positive. $$\frac{\sin \left( -44{{}^\circ}\right) \sin \left( -43{{}^\circ}\right) \cdots \sin \left( -2{{}^\circ}\right) \sin \left( -1{{}^\circ}\right) }{\sin 1{{}^\circ}\sin 2{{}^\circ}\cdots \sin 43{{}^\circ}\sin 44{{}^\circ}}=1$$ –  Américo Tavares Dec 29 '11 at 23:02
... or $$\frac{(-1)^{44}\sin 44{{}^\circ}\sin 43{{}^\circ}\cdots \sin 2{{}^\circ}\sin 1{{}^\circ}}{\sin 1{{}^\circ}\sin 2{{}^\circ}\cdots \sin 43{{}^\circ}\sin 44{{}^\circ}}=1$$ –  Américo Tavares Dec 29 '11 at 23:09
@AméricoTavares This confusion arises at all because OP does this thing called Change of variable, which I don't personally see a reason to do. But, however, thanks for dropping by your comments. –  user21436 Dec 29 '11 at 23:28

Assuming that you're asking for criticism of your calculuation: What you despearately need is some English prose connecting your various equations.

Mathematics is not supposed to be a guessing game -- but that's what results when you just dump a string of symbols such as "$x\rightarrow 45°-x$" on the unsuspecting reader and expect him to figure out for himself what it has to do with the stuff above and below it.

Instead, write it as ordinary text:

bla bla bla ... $$\prod_{x=1}^{44}\left(1-\frac{1}{\tan(x°)}\right)=\prod_{x=1}^{44} \frac{\sqrt{2}\sin(x-45°)}{\sin(x)}=2^{22} \frac{\prod_{x=1}^{44} \sin(x-45°)}{\prod_{x=1}^{44} \sin(x)}$$ Now change the index in the numerator by substituting $y=45-x$ to get $$2^{22}\frac{\prod_{y=1}^{44} -\sin(y)}{\prod_{x=1}^{44} \sin(x)}=2^{22} \prod_{x=1}^{44}\frac{-\sin x}{\sin x}$$ ... bla bla bla

That would make the argument much more readable.

As a rule of thumb, never let two displayed equation follow immediately after each other without giving the reader some actual English words to connect them. In the rare case that letting a series of equations follow directly upon each other is indeed the most readable presentation of an argument, there must be an entire, fully grammatical sentence that explains exactly what the relation between the list of equations is.

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Unless you're Fermat, right? –  mathmath8128 Dec 29 '11 at 23:55
Since I'm not Fermat, it is also true if I am Fermat. –  Henning Makholm Dec 29 '11 at 23:57
Well played... :) –  mathmath8128 Dec 29 '11 at 23:58