Question

suppose that \\( \sin a = \frac { 2 } { \sqrt { 5 } } \\) and \\( \frac { \pi } { 2 } < a < \pi \\).
find the exact values of \\( \cos \frac { a } { 2 } \\) and \\( \tan \frac { a } { 2 } \\).
\\( \cos \frac { a } { 2 } = \\)
\\( \tan \frac { a } { 2 } = \\)

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:

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