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\[ \Rightarrow \quad \cos\left(\frac{\pi}{2} + \phi\right) = 0 \]
\[ \frac{\pi}{2} + \phi = \frac{\pi}{2} + n\pi \quad \Rightarrow \quad \phi = n\pi \]
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:

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