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Home » So $2^9 \equiv 12 \pmod25$. - American Beagle Club

So $2^9 \equiv 12 \pmod25$. - American Beagle Club

📅 February 27, 2026 👤 scraface

Feb 27, 2026

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\[ \Rightarrow \quad \cos\left(\frac{\pi}{2} + \phi\right) = 0 \]
\[ \frac{\pi}{2} + \phi = \frac{\pi}{2} + n\pi \quad \Rightarrow \quad...
Now combine with \( \cos^2(\phi) = \frac{1}{2} \). The only solution s...
\[ \frac{\pi}{2} + \phi = \frac{\pi}{2} \Rightarrow \phi = 0 \] — cont...
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...
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