0
$\begingroup$

In Stein's "Complex analysis" he uses in the proof of the reflection formula (p. 164 in https://www.fing.edu.uy/~cerminar/Complex_Analysis.pdf ) the following equality:

For $0 we have $$\frac{\pi}{\sin \pi(1-s)}=\frac{\pi}{\sin\pi s}.$$ How can this be true? We have $\sin(\pi (1-s))=\sin(-\pi s)=-\sin(\pi s)$. If this equation holds for $0 then it should on all of $\mathbb{R}$. How is this not a contradiction? Where am I wrong ? Or is it the textbook?

  • 0
    If you try with $s=\frac{1}{2}$, what is happening ? Why ? and with $s=\frac{3}{2}$ ? – 2017-01-19
  • 2
    $\sin(\pi-\alpha)=\sin\alpha$ is a very basic formula of trigonometry. Apply it for $\alpha=\pi s$. – 2017-01-19

4 Answers 4

0

Just look at the sine curve on the domain $[0, \pi]$. It has a vertical line of symmetry about $s = \frac{\pi}{2}$. You have mixed up your trig identities: it is true that cos$\pi(1-s) = -$cos$\pi s$ for $s \in [0, 1]$.

3

Hint: $$\sin(\pi(1-s))=\sin(\pi-\pi s)=\sin(\pi)\cos(\pi s)-\sin(\pi s)\cos(\pi)=0 \cos(\pi s)-\sin(\pi s)(-1)=\sin(\pi s)$$

  • 0
    I don't think you can argue like this... one needs that $0 – 2017-01-19
  • 0
    @Rüdiger These are standard addition theorems for the sine function, they are true for all values of $s$ (http://mathworld.wolfram.com/TrigonometricAdditionFormulas.html). – 2017-01-20
  • 0
    you are right! sorry. +1 – 2017-01-20
3

The textbook is correct and you have a mistake on your analysis. In fact, the equality is rather easy to show:

$\sin(\pi(1-s)) = \sin(\pi - \pi s)$ and using the trivial fact that $\sin(\pi - \alpha)=\sin(\alpha)$, you have $\sin(\pi - \pi s) = \sin(\pi s)$ as argued by the author.

Moreover, the equality holds for any $s$ in the domain of the expression, but beware that any $s \in \mathbb{Z}$, and in particular $s=0$ and $s=1$, are not in the domain of $\frac{\pi}{\sin(\pi s)}$ and so you should not say that the equality holds for any real $s$.

0

$\sin(\pi - \pi s) = \sin (\pi s) $ because of the fact that sine is positive in second quadrant.