Proving, that $\text{Arg}(-i\sin(x))=\pi/2\text{sgn}(x)$ on $(-\pi,\pi)$

Question

Alright. I thought, that $\text{Arg}(-i\sin(x))=3\pi/2$, however, the Wolfram Alpha tells a different story. I am sure that it must be kind of true, because $\text{Arg}(\sin(x))$ is the result of sum of $\sum_{n=0}^\infty\frac{\sin((2n+1)x)}{2n+1}$, which is a fourier series of the sign function. But I have a little problem proving this fact.

Answer 1

As noted in the comments there are some mistakes in the question, and I don't understand what the Fourier series have to do with the problem.

Anyway, we have: $$ -\pi<x<0 \Rightarrow \sin x<0 \Rightarrow -i\sin x =i |\sin x| \Rightarrow \mbox{arg}(i |\sin x|)=\frac{\pi}{2} $$ $$ x=0 \Rightarrow \sin x=0 \Rightarrow -i\sin x =0 \Rightarrow \mbox{arg}(-i \sin x)=0 $$ $$ 0<x<\pi \Rightarrow \sin x>0 \Rightarrow -i\sin x =-i |\sin x| \Rightarrow \mbox{arg}(-i |\sin x|)=-\frac{\pi}{2} $$

so the correct result is: $ \mbox{arg}(-i \sin x)=-\frac{\pi}{2}\mbox{sgn}(x) $ for $-\pi <x<\pi$.

Answer 2

This is true $\text{Arg}(-i\sin(x))=\frac{\pi}{2}\cdot \text{sgn}(x)$, $x\in(-\pi,\pi)$

Recall that the principal argument $\text{Arg}(z)$ lies in the domain $-\pi<\text{Arg}(z)\le \pi$. Setting $y=\sin x$, then $y \in [-1,1]$. Note that $z=-iy$ is a pure imaginary complex number, and the sign of the imaginary part is $\pm$ (liying on the $y$-axis) depends on the sign of the number $y$. Thus, $\text{Arg}(-iy)=\pm\frac{\pi}{2}=\frac{\pi}{2}\text{sgn}(x)$.