Trigonometry Exact Value using Half Angle Identity I have a quick question regarding a little issue.
So I'm given a problem that says "$\tan \left(\frac{9\pi}{8}\right)$" and I'm supposed to find the exact value using half angle identities. I know what these identities are $\sin, \cos, \tan$. So, I use the tangent half-angle identity and plug-in $\theta = \frac{9\pi}{8}$ into $\frac{\theta}{2}$. I got $\frac{9\pi}{4}$ and plugged in values into the formula based on this answer. However, I checked my work with slader.com and it said I was wrong. It said I should take the value I found, $\frac{9\pi}{4}$, and plug it back into $\frac{\theta}{2}$. Wouldn't that be re-plugging in the value for no reason? Very confused.

Because the period of tangent is $\pi$, 
$\tan \dfrac{9 \pi}{8} = \tan \dfrac{\pi}{8}$
You could just look this up, but its pretty easy to derive.
$$  \tan \frac x2 
  = \frac{\sin \frac x2}{\cos \frac x2}
  = \frac{2 \sin \frac x2 \ \cos \frac x2}{1 + 2\cos^2 \frac x2 - 1}
  = \frac{\sin x}{1 + \cos x}
  = \frac{1 - \cos x}{\sin x}$$
Since $\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{1}{\sqrt 2}$
$$  \tan \dfrac{9 \pi}{8} 
  = \tan \frac{\pi}{8}
  = \tan \left( \frac 12 \cdot \frac{\pi}{4} \right)
  = \frac{1 - \cos \frac{\pi}{4}}{\sin \frac{\pi}{4}}
  = \frac{1-\frac{1}{\sqrt 2}}{\frac{1}{\sqrt 2}}
  = \sqrt 2 - 1$$
