suppose that \\( \\tan \\theta = - \\frac { 12 } { 5 } \\) and \\( 90 ^ { \\circ } < \\theta < 180 ^ {…

Explanation:

Step1: Find (\cos\theta)

Given (\tan\theta =-\frac{12}{5}) and (90^{\circ}<\theta < 180^{\circ}). We know (\tan\theta=\frac{\sin\theta}{\cos\theta}) and (\sin^{2}\theta+\cos^{2}\theta = 1). Since (\tan\theta=-\frac{12}{5}=\frac{\sin\theta}{\cos\theta}), then (\sin\theta=-\frac{12}{5}\cos\theta). Substitute into (\sin^{2}\theta+\cos^{2}\theta = 1): (\left(-\frac{12}{5}\cos\theta\right)^{2}+\cos^{2}\theta=1) (\frac{144}{25}\cos^{2}\theta+\cos^{2}\theta = 1) (\frac{144 + 25}{25}\cos^{2}\theta=1) (\cos^{2}\theta=\frac{25}{169}) Because (90^{\circ}<\theta < 180^{\circ}), (\cos\theta=-\frac{5}{13}).

Step2: Find (\sin\frac{\theta}{2})

Use the half - angle formula (\sin\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{2}}). Substitute (\cos\theta =-\frac{5}{13}) into the formula: (\sin\frac{\theta}{2}=\sqrt{\frac{1-\left(-\frac{5}{13}\right)}{2}}=\sqrt{\frac{1+\frac{5}{13}}{2}}=\sqrt{\frac{\frac{13 + 5}{13}}{2}}=\sqrt{\frac{18}{26}}=\sqrt{\frac{9}{13}}=\frac{3}{\sqrt{13}}=\frac{3\sqrt{13}}{13}).

Step3: Find (\tan\frac{\theta}{2})

Use the half - angle formula (\tan\frac{\theta}{2}=\frac{\sin\theta}{1 + \cos\theta}). First, find (\sin\theta) using (\sin^{2}\theta+\cos^{2}\theta = 1) and (\cos\theta=-\frac{5}{13}), so (\sin\theta=\frac{12}{13}). Then (\tan\frac{\theta}{2}=\frac{\frac{12}{13}}{1-\frac{5}{13}}=\frac{\frac{12}{13}}{\frac{8}{13}}=\frac{12}{8}=\frac{3}{2}).

Answer:

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