π\pi is not transcendental? Say what?

Logic Level 3

Which step did I commit a fallacy?

Step 1:

With my calculator, I get sin(201725)=1=sin(π2) \sin \left (2017 \sqrt[5]{2} \right) = -1 = \sin \left (-\frac {\pi}{2} \right )

Step 2:

Because the f(x)=sin(x)f(x) =\sin(x) has period 2π2\pi ,

201725=π2+2πn2017 \sqrt[5]{2} = -\frac {\pi}{2} + 2\pi n , for a certain integer nn

Step 3:

Rearrange the equation and set π\pi as the subject

π=201726/54n1 \LARGE \pi = \frac {2017 \cdot 2^{6/5} }{4n - 1}

This means π\pi is not a transcendental number but an algebraic number.

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