# \(\pi\) is not transcendental? Say what?

**Logic**Level 3

**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.