a) Mamy dane $\sin\alpha = \frac{12}{13}$ i $\alpha \in \left(\frac{\pi}{2}, \pi\right)$. Kąt $\alpha$ znajduje się w drugiej ćwiartce, gdzie $\cos\alpha < 0$. Najpierw obliczamy $\cos\alpha$ z jedynki trygonometrycznej:\[\sin^2\alpha + \cos^2\alpha = 1\]\[\left(\frac{12}{13}\right)^2 + \cos^2\alpha = 1\]\[\frac{144}{169} + \cos^2\alpha = 1\]\[\cos^2\alpha = 1 - \frac{144}{169} = \frac{25}{169}\]\[\cos\alpha = -\sqrt{\frac{25}{169}} = -\frac{5}{13}\]Teraz obliczamy wartości funkcji trygonometrycznych dla kąta $2\alpha$:\[\sin 2\alpha = 2\sin\alpha\cos\alpha = 2 \cdot \frac{12}{13} \cdot \left(-\frac{5}{13}\right) = -\frac{120}{169}\]\[\cos 2\alpha = \cos^2\alpha - \sin^2\alpha = \left(-\frac{5}{13}\right)^2 - \left(\frac{12}{13}\right)^2 = \frac{25}{169} - \frac{144}{169} = -\frac{119}{169}\]\[\operatorname{tg} 2\alpha = \frac{\sin 2\alpha}{\cos 2\alpha} = \frac{-120/169}{-119/169} = \frac{120}{119}\]\[\operatorname{ctg} 2\alpha = \frac{1}{\operatorname{tg} 2\alpha} = \frac{119}{120}\]b) Mamy dane $\cos\alpha = \frac{5}{13}$ i $\alpha \in \left(\frac{3\pi}{2}, 2\pi\right)$. Kąt $\alpha$ znajduje się w czwartej ćwiartce, gdzie $\sin\alpha < 0$. Najpierw obliczamy $\sin\alpha$ z jedynki trygonometrycznej:\[\sin^2\alpha + \cos^2\alpha = 1\]\[\sin^2\alpha + \left(\frac{5}{13}\right)^2 = 1\]\[\sin^2\alpha + \frac{25}{169} = 1\]\[\sin^2\alpha = 1 - \frac{25}{169} = \frac{144}{169}\]\[\sin\alpha = -\sqrt{\frac{144}{169}} = -\frac{12}{13}\]Teraz obliczamy wartości funkcji trygonometrycznych dla kąta $2\alpha$:\[\sin 2\alpha = 2\sin\alpha\cos\alpha = 2 \cdot \left(-\frac{12}{13}\right) \cdot \frac{5}{13} = -\frac{120}{169}\]\[\cos 2\alpha = \cos^2\alpha - \sin^2\alpha = \left(\frac{5}{13}\right)^2 - \left(-\frac{12}{13}\right)^2 = \frac{25}{169} - \frac{144}{169} = -\frac{119}{169}\]\[\operatorname{tg} 2\alpha = \frac{\sin 2\alpha}{\cos 2\alpha} = \frac{\frac{-120}{169}}{\frac{-119}{169}} = \frac{120}{119}\]\[\operatorname{ctg} 2\alpha = \frac{1}{\operatorname{tg} 2\alpha} = \frac{119}{120}\]c) Mamy dane $\operatorname{tg}\alpha = 3$ i $\alpha \in \left(0, \frac{\pi}{2}\right)$.\[\operatorname{tg} 2\alpha = \frac{2\operatorname{tg}\alpha}{1 - \operatorname{tg}^2\alpha} = \frac{2 \cdot 3}{1 - 3^2} = \frac{6}{1 - 9} = -\frac{3}{4}\]\[\operatorname{ctg} 2\alpha = -\frac{4}{3}\]Znajdujemy $\sin\alpha$ i $\cos\alpha$. Skoro $\operatorname{tg}\alpha=3$ i $\alpha$ jest w I ćwiartce, to $\sin\alpha = \frac{3}{\sqrt{10}}$ i $\cos\alpha = \frac{1}{\sqrt{10}}$.\[\sin 2\alpha = 2\sin\alpha\cos\alpha = 2 \cdot \frac{3}{\sqrt{10}} \cdot \frac{1}{\sqrt{10}} = \frac{6}{10} = \frac{3}{5}\]\[\cos 2\alpha = \cos^2\alpha - \sin^2\alpha = \frac{1}{10} - \frac{9}{10} = -\frac{8}{10} = -\frac{4}{5}\]d) Mamy dane $\operatorname{ctg}\alpha = -\frac{3}{4}$ i $\alpha \in \left(\frac{\pi}{2}, \pi\right)$. Zatem $\operatorname{tg}\alpha = -\frac{4}{3}$.\[\operatorname{tg} 2\alpha = \frac{2\operatorname{tg}\alpha}{1 - \operatorname{tg}^2\alpha} = \frac{2 \cdot \left(-\frac{4}{3}\right)}{1 - \left(-\frac{4}{3}\right)^2} = \frac{\frac{-8}{3}}{1 - \frac{16}{9}} = \frac{\frac{-8}{-3}}{\frac{-7}{9}} = \frac{24}{7}\]\[\operatorname{ctg} 2\alpha = \frac{7}{24}\]Znajdujemy $\sin\alpha$ i $\cos\alpha$. Skoro $\operatorname{ctg}\alpha=-\frac{3}{4}$ i $\alpha$ jest w II ćwiartce, to $\sin\alpha = \frac{4}{5}$ i $\cos\alpha = -\frac{3}{5}$.\[\sin 2\alpha = 2\sin\alpha\cos\alpha = 2 \cdot \frac{4}{5} \cdot \left(-\frac{3}{5}\right) = -\frac{24}{25}\]\[\cos 2\alpha = \cos^2\alpha - \sin^2\alpha = \left(-\frac{3}{5}\right)^2 - \left(\frac{4}{5}\right)^2 = \frac{9}{25} - \frac{16}{25} = -\frac{7}{25}\]