422. Pokažite, da velja: a) \( \cos \left(\frac{\pi}{2}+\alpha\right) \cos \left(\frac{\pi}{2}-\alpha\right)+\sin ^{2}(\alpha+2 \pi)=0 \)
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Da bi dokazali, da velja navedena enačba, najprej analiziramo posamezne dele. Uporabimo trigonometrične identitete: 1. **Prvi del**: \[ \cos \left( \frac{\pi}{2} + \alpha \right) = -\sin(\alpha) \quad \text{in} \quad \cos \left( \frac{\pi}{2} - \alpha \right) = \sin(\alpha) \] Tako dobimo: \[ \cos \left( \frac{\pi}{2} + \alpha \right) \cos \left( \frac{\pi}{2} - \alpha \right) = (-\sin(\alpha))(\sin(\alpha)) = -\sin^2(\alpha) \] 2. **Drugi del**: \[ \sin^2(\alpha + 2\pi) = \sin^2(\alpha) \quad \text{(ker je faktor } 2\pi \text{ enak nobeni rotaciji)} \] Sedaj združimo rezultate: \[ -\sin^2(\alpha) + \sin^2(\alpha) = 0 \] Tako dobimo, da velja: \[ \cos \left(\frac{\pi}{2}+\alpha\right) \cos \left(\frac{\pi}{2}-\alpha\right)+\sin^{2}(\alpha+2\pi)=0 \] Dokazano!