Вычисления и преобразования

$$\sin^2 \alpha + \cos^2 \alpha = 1 \implies \cos^2 \alpha = 1 - \left(-\frac{2\sqrt{2}}{3}\right)^2 = \frac{1}{9}$$ $$\cos \alpha = \frac{1}{3} \quad \alpha \in (IV)$$ $$3\cos \alpha = 1$$

$$\sin^2 \alpha = 1 - \left(\frac{2\sqrt{6}}{5}\right)^2 = 1 - \frac{24}{25} = \frac{1}{25}$$ $$\sin \alpha = -\frac{1}{5} \quad (\alpha \in \text{IV четверти})$$ $$5\sin \alpha = -1$$

$$\cos 2\alpha = 1 - 2\sin^2 \alpha = 1 - 2 \cdot (0,2)^2 = 1 - 0,08 = 0,92$$ $$24\cos 2\alpha = 24 \cdot 0,92 = 22,08$$

$$\sin 6\alpha = 2\sin 3\alpha \cos 3\alpha = 2 \cdot 0,6 \cdot \cos 3\alpha$$ $$\frac{10\sin 6\alpha }{3\cos 3\alpha } = \frac{10 \cdot 2 \cdot 0,6 \cdot \cos 3\alpha}{3\cos 3\alpha} = \frac{12}{3} = 4$$

$$\cos (\pi - \beta) = -\cos \beta, \quad \sin \left(\frac{\pi}{2} + \beta\right) = \cos \beta$$ $$\cos (\beta + 3\pi) = \cos (\beta + \pi + 2\pi) = \cos (\beta + \pi) = -\cos \beta$$ $$\frac{3(-\cos \beta) + \cos \beta}{-\cos \beta} = \frac{-2\cos \beta}{-\cos \beta} = 2$$