if secα = 17/8, 3π/2 < α < 2π, then find the exact value of each of the following.\na. sin α/2\nb. cos…

Explanation:

Step1: Find (\cos\alpha)

Given (\sec\alpha=\frac{17}{8}), and (\sec\alpha=\frac{1}{\cos\alpha}), so (\cos\alpha = \frac{8}{17}). Since (\frac{3\pi}{2}<\alpha < 2\pi), (\sin\alpha=-\sqrt{1 - \cos^{2}\alpha}=-\sqrt{1-\left(\frac{8}{17}\right)^{2}}=-\sqrt{\frac{289 - 64}{289}}=-\frac{15}{17}).

Step2: Calculate (\sin\frac{\alpha}{2})

Use the half - angle formula (\sin\frac{\alpha}{2}=\pm\sqrt{\frac{1-\cos\alpha}{2}}). Because (\frac{3\pi}{4}<\frac{\alpha}{2}<\pi) (divide (\frac{3\pi}{2}<\alpha < 2\pi) by 2), (\sin\frac{\alpha}{2}>0). (\sin\frac{\alpha}{2}=\sqrt{\frac{1-\frac{8}{17}}{2}}=\sqrt{\frac{\frac{9}{17}}{2}}=\sqrt{\frac{9}{34}}=\frac{3}{\sqrt{34}}=\frac{3\sqrt{34}}{34}).

Step3: Calculate (\cos\frac{\alpha}{2})

Use the half - angle formula (\cos\frac{\alpha}{2}=\pm\sqrt{\frac{1 + \cos\alpha}{2}}). Since (\frac{3\pi}{4}<\frac{\alpha}{2}<\pi), (\cos\frac{\alpha}{2}<0). (\cos\frac{\alpha}{2}=-\sqrt{\frac{1+\frac{8}{17}}{2}}=-\sqrt{\frac{\frac{25}{17}}{2}}=-\sqrt{\frac{25}{34}}=-\frac{5}{\sqrt{34}}=-\frac{5\sqrt{34}}{34}).

Step4: Calculate (\tan\frac{\alpha}{2})

Use the formula (\tan\frac{\alpha}{2}=\frac{\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}}) or (\tan\frac{\alpha}{2}=\frac{1-\cos\alpha}{\sin\alpha}). Using (\tan\frac{\alpha}{2}=\frac{1-\cos\alpha}{\sin\alpha}), substitute (\cos\alpha=\frac{8}{17}) and (\sin\alpha =-\frac{15}{17}). (\tan\frac{\alpha}{2}=\frac{1-\frac{8}{17}}{-\frac{15}{17}}=\frac{\frac{9}{17}}{-\frac{15}{17}}=-\frac{3}{5}).

Answer:

a. (\frac{3\sqrt{34}}{34}) b. (-\frac{5\sqrt{34}}{34}) c. (-\frac{3}{5})