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.

  • $\begingroup$ Welcome to MSE. Please use math.meta.stackexchange.com/questions/5020/… in writing equations so that it is understandable and attractive to look at.... $\endgroup$ – fitzmerl duron Feb 18 '18 at 3:30
  • $\begingroup$ It's not clear what you mean by when you say you plug something into $\frac\theta2.$ Note that $\frac{9\pi/8}{2} = \frac{9\pi}{16},$ so if what you meant is that you replaced each $\theta$ with $\frac{9\pi}8,$ you made an arithmetic error. $\endgroup$ – David K Feb 18 '18 at 3:54

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 $x=\tan \frac98\pi$


which gives

$$y=\tan \frac98\pi=\sqrt 2 -1$$

as acceptable answer.

  • $\begingroup$ Are you sure about y= tan 9π/4 . I guess it should have been tan 9π/8. Or else your quadratic must have been x²+2x-1=0 according to your notation $\endgroup$ – Rohan Shinde Feb 18 '18 at 3:53
  • $\begingroup$ Ops just a typo, I fix thanks! $\endgroup$ – user Feb 18 '18 at 3:55
  • $\begingroup$ First you stated tan 9π/4=1 and afterwards you find tan 9π/4=√2-1 $\endgroup$ – Rohan Shinde Feb 18 '18 at 3:55
  • $\begingroup$ Now it's fixed. $\endgroup$ – Rohan Shinde Feb 18 '18 at 3:59

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.


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 $\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{1}{\sqrt 2}$

$$ \tan \dfrac{9 \pi}{8} = \tan \frac{\pi}{8} = \tan \left( \frac 12 \cdot \frac{\pi}{4} \right) = \frac{1 - \cos \frac{\pi}{4}}{\sin \frac{\pi}{4}} = \frac{1-\frac{1}{\sqrt 2}}{\frac{1}{\sqrt 2}} = \sqrt 2 - 1$$


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