Answer:
[tex]= \sqrt{\frac{1 - \frac{\sqrt{2 + \sqrt{2}}}{2}}{1 + \frac{\sqrt{2 + \sqrt{2}}}{2}}} \][/tex]
Step-by-step explanation:
[tex]To express \( \tan(\frac{\pi}{16}) \) using the half-angle identity, we can use the formula:\[ \tan\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos(\theta)}{1 + \cos(\theta)}} \]where \( \theta = \frac{\pi}{8} \), since \( \frac{\pi}{16} = \frac{\frac{\pi}{8}}{2} \).Now, let's compute:\[ \cos\left(\frac{\pi}{8}\right) = \frac{\sqrt{2 + \sqrt{2}}}{2} \]Then,\[ \tan\left(\frac{\pi}{16}\right) = \sqrt{\frac{1 - \cos\left(\frac{\pi}{8}\right)}{1 + \cos\left(\frac{\pi}{8}\right)}} \][/tex][tex]= \sqrt{\frac{1 - \cos\left(\frac{\pi}{8}\right)}{1 + \cos\left(\frac{\pi}{8}\right)}} \]\[ = \sqrt{\frac{1 - \frac{\sqrt{2 + \sqrt{2}}}{2}}{1 + \frac{\sqrt{2 + \sqrt{2}}}{2}}} \]This is the exact expression for \( \tan\left(\frac{\pi}{16}\right) \) using the half-angle identity.[/tex]