5. Natančno izračunaj \( \sin \left(\alpha+\frac{5 \pi}{6}\right) \), če je \( \sin \alpha=-\frac{4}{5} \) in \( \pi<\alpha<\frac{3 \pi}{2} \).

Da natančno izračunamo \( \sin \left(\alpha + \frac{5 \pi}{6}\right) \), bomo uporabili formulo za vsoto sinusov: \[ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \] V našem primeru je \( \beta = \frac{5 \pi}{6} \). Če najprej izračunamo \( \cos \beta \) in \( \sin \beta \): - \( \sin \left(\frac{5 \pi}{6}\right) = \sin \left(\pi - \frac{\pi}{6}\right) = \sin \left(\frac{\pi}{6}\right) = \frac{1}{2} \) - \( \cos \left(\frac{5 \pi}{6}\right) = -\cos \left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2} \) Ker imamo \( \sin \alpha = -\frac{4}{5} \) in \( \pi < \alpha < \frac{3\pi}{2} \), lahko izračunamo \( \cos \alpha \) s pomočjo Pitagorovega izreka: \[ \sin^2 \alpha + \cos^2 \alpha = 1 \implies \cos^2 \alpha = 1 - \left(-\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25} \] Torej je \( \cos \alpha = -\sqrt{\frac{9}{25}} = -\frac{3}{5} \) (negativen, ker je v tretjem kvadrantu). Sedaj lahko vstavimo vrednosti v formulo: \[ \sin \left(\alpha + \frac{5 \pi}{6}\right) = \sin \alpha \cos \left(\frac{5 \pi}{6}\right) + \cos \alpha \sin \left(\frac{5 \pi}{6}\right) \] \[ = \left(-\frac{4}{5}\right) \left(-\frac{\sqrt{3}}{2}\right) + \left(-\frac{3}{5}\right) \left(\frac{1}{2}\right) \] \[ = \frac{4\sqrt{3}}{10} - \frac{3}{10} = \frac{4\sqrt{3} - 3}{10} \] Tako dobimo: \[ \sin \left(\alpha + \frac{5 \pi}{6}\right) = \frac{4\sqrt{3} - 3}{10} \]