suppose that \\( \\tan \\alpha=-\\frac{12}{5} \\) and \\( 270^{\\circ}<\\alpha<360^{\\circ} \\).\nfind the…

Explanation:

Step1: Find (\cos\alpha)

Since (\tan\alpha =-\frac{12}{5}=\frac{\sin\alpha}{\cos\alpha}) and (\sin^{2}\alpha+\cos^{2}\alpha = 1), then (\left(-\frac{12}{5}\cos\alpha\right)^{2}+\cos^{2}\alpha=1). (\frac{144}{25}\cos^{2}\alpha+\cos^{2}\alpha = 1), (\frac{144 + 25}{25}\cos^{2}\alpha=1), (\cos^{2}\alpha=\frac{25}{169}). Because (270^{\circ}<\alpha<360^{\circ}), (\cos\alpha=\frac{5}{13}).

Step2: Use the half - angle formula for (\sin\frac{\alpha}{2})

The half - angle formula (\sin\frac{\alpha}{2}=\pm\sqrt{\frac{1-\cos\alpha}{2}}). Since (135^{\circ}<\frac{\alpha}{2}<180^{\circ}), (\sin\frac{\alpha}{2}>0). (\sin\frac{\alpha}{2}=\sqrt{\frac{1-\frac{5}{13}}{2}}=\sqrt{\frac{\frac{8}{13}}{2}}=\sqrt{\frac{4}{13}}=\frac{2}{\sqrt{13}}=\frac{2\sqrt{13}}{13}).

Step3: Use the half - angle formula for (\tan\frac{\alpha}{2})

The half - angle formula (\tan\frac{\alpha}{2}=\frac{\sin\alpha}{1 + \cos\alpha}). Since (\tan\alpha=-\frac{12}{5}=\frac{\sin\alpha}{\cos\alpha}) and (\cos\alpha=\frac{5}{13}), then (\sin\alpha=-\frac{12}{13}). (\tan\frac{\alpha}{2}=\frac{-\frac{12}{13}}{1+\frac{5}{13}}=\frac{-\frac{12}{13}}{\frac{18}{13}}=-\frac{2}{3}).

Answer:

(\sin\frac{\alpha}{2}=\frac{2\sqrt{13}}{13}), (\tan\frac{\alpha}{2}=-\frac{2}{3})