# How do you use the half-angle formulas to find the exact value of tan((3pi)/8)?

##### 2 Answers
Jun 14, 2015

Find $\tan \left(\frac{3 \pi}{8}\right)$

#### Explanation:

Use trig identity: $\tan 2 a = \frac{2 \tan a}{1 - {\tan}^{2} a}$

Call $\tan \left(\frac{3 \pi}{8}\right) = t$ . Trig table gives -->$\tan \left(\frac{3 \pi}{4}\right) = - 1$

$\frac{2 t}{1 - {t}^{2}} = - 1 \to 2 t = {t}^{2} - 1$ -> Solve quadratic equation:

${t}^{2} - 2 t - 1 = 0$

$D = {d}^{2} = 4 + 4 = 8 \to d = \pm 2 \sqrt{2}$

$t = \tan \left(\frac{3 \pi}{8}\right) = 1 + \left(\sqrt{2}\right)$ (Quadrant I)

Jun 15, 2015

$\tan \left(\frac{3 \pi}{8}\right) = \tan \left(\frac{3 \pi}{4} / 2\right)$

To get the half-angle identity, recall that:

$\sin \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 - \cos x}{2}}$
$+$ in quadrants I and II
$-$ in quadrants III and IV

$\cos \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 + \cos x}{2}}$
$+$ in quadrants I and IV
$-$ in quadrants II and III

You can get the quadrant conditions by saying that $y = \sin x$ is above the x-axis on the first-two ${90}^{o}$-intervals and below the x-axis on the second-two ${90}^{o}$-intervals. Similar for $y = \cos x$.

Thus:

$\tan \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 - \cos x}{1 + \cos x}}$
$+$ in quadrants I and III
$-$ in quadrants II and IV
(you can get these quadrant conditions after dividing the quadrant conditions of $\sin$ and $\cos$.)

$\frac{3 \pi}{8} \cdot \left({180}^{o} / \pi\right)$ is ${67.5}^{o}$ and so it's in quadrant I (${67.5}^{o} < {90.0}^{o}$), making the condition $+$.

$\tan \left(\frac{3 \pi}{8}\right) = \sqrt{\frac{1 - \cos \left(\frac{3 \pi}{4}\right)}{1 + \cos \left(\frac{3 \pi}{4}\right)}}$

Now we need to figure out $\cos \left(\frac{3 \pi}{4}\right)$. $\cos x$ in $\frac{\pi}{4}$ intervals has the pattern $1 , \frac{\sqrt{2}}{2} , 0 , - \frac{\sqrt{2}}{2} , - 1 , - \frac{\sqrt{2}}{2} , 0 , \frac{\sqrt{2}}{2} , 1 , \ldots$, with $\cos \left(\frac{0 \pi}{4}\right) = 1$.

Thus:
$\cos \left(\frac{3 \pi}{4}\right) = - \frac{\sqrt{2}}{2}$

=> sqrt((1+sqrt2/2)/(1-sqrt2/2)

Get common denominators of 2:
= sqrt(((2+sqrt2)/2)/((2-sqrt2)/2)

Cancel:
$= \sqrt{\frac{2 + \sqrt{2}}{2 - \sqrt{2}}}$

Now let's just make it look nicer. Multiply by the conjugate of the bottom $\left(\sqrt{2 + \sqrt{2}}\right)$:

$= {\left(\sqrt{2 + \sqrt{2}}\right)}^{2} / \sqrt{4 - {\left(\sqrt{2}\right)}^{2}}$

$= \frac{2 + \sqrt{2}}{\sqrt{2}}$

Split and simplify by multiplying by "$1$", cancel things out:
$= 1 + \frac{2}{\sqrt{2}} = 1 \cdot 1 + \frac{2 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = 1 \cdot 1 + \frac{2 \sqrt{2}}{2} = 1 + \sqrt{2}$