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.
 A: You need to use 
$$\tan 2x=\frac{2\tan x}{1-\tan^2x}$$
with $$x=\frac98\pi$$
and since we know that $$\tan 2x=\tan \frac94\pi=\tan \frac{\pi}4=1$$
we have with $x=\tan \frac98\pi$
$$x^2+2x-1=0$$
which gives 
$$y=\tan \frac98\pi=\sqrt 2 -1$$
as acceptable answer.
A: Let $\tan \frac {9\pi}{8}= \tan \frac {\theta }{2}=a$
By half angle formulas $$\tan \theta=\frac {2\tan \frac {\theta }{2}}{1-\tan ^2\frac {\theta }{2}}$$
Hence we get $$1=\frac {2a}{1-a^2}$$
Hence we get $a^2+2a-1=0$
Solve the quadratic to get the answer.
Note : By using quadratic formula we get $\tan \frac {9\pi}{8}= \sqrt 2 -1$
The other solution i.e. $-\sqrt 2-1$ gets rejected because $\frac {9\pi}{8}$ lies in third quadrant where $\tan $ must be positive.
A: 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 $\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{1}{\sqrt 2}$
$$  \tan \dfrac{9 \pi}{8} 
  = \tan \frac{\pi}{8}
  = \tan \left( \frac 12 \cdot \frac{\pi}{4} \right)
  = \frac{1 - \cos \frac{\pi}{4}}{\sin \frac{\pi}{4}}
  = \frac{1-\frac{1}{\sqrt 2}}{\frac{1}{\sqrt 2}}
  = \sqrt 2 - 1$$
