# $$\pi$$ is not transcendental? Say what?

Logic Level 3

Which step did I commit a fallacy?

Step 1:

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

Step 2:

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

$$2017 \sqrt[5]{2} = -\frac {\pi}{2} + 2\pi n$$, for a certain integer $$n$$

Step 3:

Rearrange the equation and set $$\pi$$ as the subject

$$\LARGE \pi = \frac {2017 \cdot 2^{6/5} }{4n - 1}$$

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

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