suppose that \\( \\sin a = \\frac { 2 } { \\sqrt { 5 } } \\) and \\( \\frac { \\pi } { 2 } < a < \\pi…

Explanation:

Step1: Find (\cos\alpha)

Since (\sin\alpha=\frac{2}{\sqrt{5}}) and (\frac{\pi}{2}<\alpha <\pi), using (\sin^{2}\alpha+\cos^{2}\alpha = 1), we have (\cos\alpha=-\sqrt{1-\sin^{2}\alpha}=-\sqrt{1 - (\frac{2}{\sqrt{5}})^2}=-\frac{1}{\sqrt{5}})

Step2: Find (\cos\frac{\alpha}{2})

Using the half - angle formula (\cos\frac{\alpha}{2}=\pm\sqrt{\frac{1 + \cos\alpha}{2}}). Because (\frac{\pi}{4}<\frac{\alpha}{2}<\frac{\pi}{2}) (so (\cos\frac{\alpha}{2}>0)), then (\cos\frac{\alpha}{2}=\sqrt{\frac{1-\frac{1}{\sqrt{5}}}{2}}=\sqrt{\frac{\sqrt{5}-1}{2\sqrt{5}}}=\frac{\sqrt{\sqrt{5}-1}}{\sqrt{2\sqrt{5}}}=\frac{\sqrt{5\sqrt{5}- 5}}{5})

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

Using the half - angle formula (\tan\frac{\alpha}{2}=\frac{\sin\alpha}{1+\cos\alpha}) Substitute (\sin\alpha=\frac{2}{\sqrt{5}}) and (\cos\alpha=-\frac{1}{\sqrt{5}}) into it: (\tan\frac{\alpha}{2}=\frac{\frac{2}{\sqrt{5}}}{1-\frac{1}{\sqrt{5}}}=\frac{2}{\sqrt{5}-1}=\frac{2(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}=\frac{\sqrt{5}+1}{2})

Answer:

(\cos\frac{\alpha}{2}=\frac{\sqrt{5\sqrt{5}-5}}{5}), (\tan\frac{\alpha}{2}=\frac{\sqrt{5}+1}{2})