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Home » \( \frac{360 \text miles}{12 \text gal} = 30 \text mpg \). - Aurero

\( \frac{360 \text miles}{12 \text gal} = 30 \text mpg \). - Aurero

📅 March 14, 2026 👤 scraface

Mar 14, 2026

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📌 \[ \frac{\pi}{2} + \phi = \frac{\pi}{2} + n\pi \quad \Rightarrow \quad \phi = n\pi \]

📌 Now combine with \( \cos^2(\phi) = \frac{1}{2} \). The only solution satisfying both is \( \phi = -\frac{\pi}{4} \) (since \( \phi = \frac{\pi}{4} \) would give \( \cos^2 = \frac{1}{2} \), but \( \cos\left(\frac{\pi}{2} + \frac{\pi}{4}\right) = \cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}} \), not zero). However, satisfying \( \cos\left(\frac{\pi}{2} + \phi\right) = 0 \) implies:

📌 \[ \frac{\pi}{2} + \phi = \frac{\pi}{2} \Rightarrow \phi = 0 \] — contradiction.

📌 Wait: \( \cos\left(\frac{\pi}{2} + \phi\right) = -\sin(\phi) \), so:

📌 \[ \cos^2\left(\frac{\pi}{2} + \phi\right) = \sin^2(\phi) = 0 \]

📌 \[ \sin(\phi) = 0 \]

📌 Thus \( \phi = k\pi \). Now use \( \cos^2(\phi) = \frac{1}{2} \), but \( \cos(k\pi) = \pm 1 \), so \( \cos

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