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How do we compute : $$i)\ S_1 = \tan1-\tan3+\tan5-\cdots+\tan89$$ and $$ii)\ S_2 = \tan1+\tan3+\tan5+\cdots+\tan89$$

all the angles are in degrees. Thanks

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    $\begingroup$ @YiyuanLee : that's not possible because $S_2 > \tan 89 = 1/\tan 1 \approx 180/\pi >> 97/2\pi$ $\endgroup$
    – mercio
    Apr 15, 2014 at 16:53
  • $\begingroup$ Sorry, got it mixed up. It should be: $S_2$ is numerically very close to $\frac{97\pi}{2}$. $\endgroup$
    – Yiyuan Lee
    Apr 15, 2014 at 17:04
  • $\begingroup$ Unfortunately, the second sum is not exactly $\dfrac{97\pi}{2}$, but it is certainly close. It adds up to about $48.500031953\pi$. (Good thing, because it was hard to imagine any kind of proof that would give $\dfrac{97\pi}{2}$ as an answer!) $\endgroup$
    – Steve Kass
    Apr 16, 2014 at 0:57

2 Answers 2

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Numerically, $S_1$ is very close to 45, this suggest it might have a simple derivation. Indeed, it does have one. For $S_2$, I have no idea.

For $S_1$, notice $$\begin{array}{rrrrr} \tan 1^\circ = \cot 89^\circ,& \tan 5^\circ = \cot 84^\circ,& \ldots,& \tan k^\circ = \cot (90-k)^\circ,& \ldots\\ -\tan 3^\circ = \cot 93^\circ,& -\tan 7^\circ = \cot 97^\circ,&\ldots,& -\tan k^\circ = \cot(90+k)^\circ,&\ldots \end{array}$$ We can rewrite $S_1$ as

$$S_1 = \sum_{k=0}^{44}\cot(4k+1)^\circ = \sum_{k=0}^{44}\cot\left(\frac{\pi k}{45} + \frac{\pi}{180}\right)$$

Using following product formula for sine:

$$\sin(Nx) = 2^{N-1}\prod_{k=0}^{N-1} \sin\left(\frac{\pi k}{N} + x\right)$$

take logarithm and differentiate with respect to $x$ gives us

$$N\cot(Nx) = \sum_{k=0}^{N-1} \cot\left(\frac{\pi k}{N} + x\right)$$

and hence

$$S_1 = 45\cot\left(45\times\frac{\pi}{180}\right) = 45\cot\left(\frac{\pi}{4}\right) = 45.$$

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  • $\begingroup$ Good job. You beat me to the answer by 10 seconds! $\endgroup$
    – Steve Kass
    Apr 15, 2014 at 17:43
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    $\begingroup$ @SteveKass I original thought you beat me by 10 seconds. I'm happier now ;-p $\endgroup$ Apr 15, 2014 at 17:46
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The value of i) is 45.

Mathematica computed the value numerically, and that suggested I look for a solution involving products and a derivative. Here's a derivation.

According to this post, the following is a trigonometric identity.

$$2\,\sin \left( n\theta \right) =\prod _{k=0}^{n-1}2\,\sin \left( \theta+{\frac {k\pi }{n}} \right)$$

Let $n=45$ to get the following identity.

$$2\,\sin \left( 45\theta \right) =\prod _{k=0}^{44}2\,\sin \left( \theta+k4^\circ \right)$$

Differentiate both sides with respect to $\theta$ and use the original identity to get the following identity.

$$90\,\cos\left( 45\theta \right)=\left(\prod _{k=0}^{44}2\,\sin \left( \theta+k4^\circ \right)\right)\sum_{k=0}^{44}\cot\left( \theta+k4^\circ \right) =2\sin(45\theta)\sum_{k=0}^{44}\cot\left( \theta+k4^\circ \right)$$

Let $\theta=1^\circ$ to get the following.

$$45\,\cot\left( 45^\circ \right)=\cot(1^\circ)+\cot(5^\circ)+\cdots+\cot(177^\circ)$$

Using basic trig identities,

$$45=\tan(89^\circ)+\tan(85^\circ)+\cdots+\tan(1^\circ)+\cot(177^\circ)+\cot(173^\circ)+\cdots\cot(93^\circ),$$

which with a little rearrangement is the desired result.

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