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why am i always horny - Richter Guitar

📅 May 4, 2026 👤 scraface

May 04, 2026

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📌 $\sin(90^\circ) = 1$, $\cos(90^\circ) = 0$ â $0

📌 e 1$ â Wait! But $\cos z = 0$, so only valid if $2\sin z - 1 = 0$, but $2(1) - 1 = 1

📌 e 0$. Contradiction?

📌 Wait â at $z = 90^\circ$, $\cos z = 0$, so LHS: $\sin(180^\circ) = 0$, RHS: $\cos(90^\circ) = 0$ â $0 = 0$, valid.

📌 $\sin(2 \cdot 90^\circ) = \sin(180^\circ) = 0$, $\cos(90^\circ) = 0$ â satisfies.

📌 $\sin(270^\circ) = -1$, $\cos(270^\circ) = 0$ â $-1 = 0$? No â Not valid.

📌 So only solutions where $\cos z = 0$ and $\sin z = 1/2$, but $\cos z = 0$ at $z = 90^\circ, 270^\circ$, but $\sin(270^\circ) = -1

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