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I have to study this function. To remove the modulus I split the original function in

\begin{cases} \log(1+\sin(x)+\sin(x)), & \text{if $0<x<\pi$} \\ \log(1+\sin(x)-\sin(x)), & \text{if $\pi<x<2\pi$} \end{cases}

but since $\log(1+\sin(x)-\sin(x)) = \log(1) = 0$, can I study the function only in the interval $0<x<\pi$?

colis
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1 Answers1

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Yes, it remains to study only the cases $2k\pi<x<(2k+1)\pi$ (which are obviously the same as $0<x<\pi$), as we already know that the function gives $0$ elsewhere.

Berci
  • 90,745