Question 811967
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Define *[tex \Large f\[0,2\pi\]\,\right\,y\ \in\ \mathbb{R}\ :\ f(0)\ =\ f(2\pi)]


Let *[tex \Large f(x)\ =\ \sin(x)]


*[tex \Large \left(\forall\,x\ \in\ \mathbb{R}\right)\left(y\ =\ \sin(x)\ \rightarrow \ y\ \in\ \[-1,1\]\ \in\ \mathbb{R}\right)]


and


*[tex \Large \sin(0)\ =\ 0\ \small{\wedge}\ \Large \sin(2\pi)\ =\ 0]


Show *[tex \Large \exists\,x\ \in\ \[0,\pi\] :\ f(x)\ =\ f(x\ +\ \pi)]


Let *[tex \Large x\ \in\ \{0,\pi\}]


Then *[tex \Large f(0)\ =\ 0\ \small{\wedge}\ \Large f(\pi)\ =\ 0\ \small{\wedge}\ \Large f(2\pi)\ =\ 0]


Interestingly enough, the above result notwithstanding:


*[tex \Large \left(\forall\,x\ \in\ \mathbb{R}\right)\ \left(\sin(x)\ =\ -\sin(x\ +\ \pi)\right)]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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