# How do you find the exact values of tan 112.5 degrees using the half angle formula?

Jul 22, 2015

$\tan \left(112.5\right) = - \left(1 + \sqrt{2}\right)$

#### Explanation:

$112.5 = 112 \frac{1}{2} = \frac{225}{2}$

NB : This angle lies in the 2nd Quadrant.

$\implies \tan \left(112.5\right) = \tan \left(\frac{225}{5}\right) = \sin \frac{\frac{225}{2}}{\cos} \left(\frac{225}{2}\right) = - \sqrt{{\left[\sin \frac{\frac{225}{2}}{\cos} \left(\frac{225}{2}\right)\right]}^{2}} = - \sqrt{{\sin}^{2} \frac{\frac{225}{2}}{\cos} ^ 2 \left(\frac{225}{2}\right)}$

We say it's negative because the value of $\tan$ is always negative in the second quadrant!

Next, we use the half angle formula below :

${\sin}^{2} \left(\frac{x}{2}\right) = \frac{1}{2} \left(1 - \cos x\right)$

${\cos}^{2} \left(\frac{x}{2}\right) = \frac{1}{2} \left(1 + \cos x\right)$

$\implies \tan \left(112.5\right) = - \sqrt{{\sin}^{2} \frac{\frac{225}{2}}{\cos} ^ 2 \left(\frac{225}{2}\right)} = - \sqrt{\frac{\frac{1}{2} \left(1 - \cos \left(225\right)\right)}{\frac{1}{2} \left(1 + \cos \left(225\right)\right)}} = - \sqrt{\frac{1 - \cos \left(225\right)}{1 + \cos \left(225\right)}}$

Notice that : $225 = 180 + 45 \implies \cos \left(225\right) = - \cos \left(45\right)$

$\implies \tan \left(112.5\right) = - \sqrt{\frac{1 - \left(- \cos 45\right)}{1 + \left(- \cos 45\right)}} = - \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{1 - \frac{\sqrt{2}}{2}}} = \sqrt{\frac{2 + \sqrt{2}}{2 - \sqrt{2}}}$

Now you want to Rationalize;

$\implies - \sqrt{\frac{\left(2 + \sqrt{2}\right) \times \left(2 + \sqrt{2}\right)}{\left(2 - \sqrt{2}\right) \times \left(2 + \sqrt{2}\right)}} = - \sqrt{\frac{{\left(2 + \sqrt{2}\right)}^{2}}{4 - 2}} = - \frac{2 + \sqrt{2}}{\sqrt{2}} = - \frac{\sqrt{2} \times \left(2 + \sqrt{2}\right)}{\sqrt{2} \times \sqrt{2}} = - \frac{2 \sqrt{2} + 2}{2} = \textcolor{b l u e}{- \left(1 + \sqrt{2}\right)}$

Jul 22, 2015

Find tan 112.5

Ans: (-1 - sqrt2)

#### Explanation:

Call tan 112.5 = tan t
tan 2t = tan 225 = tan (45 + 180) = tan 45 = 1
Use trig identity: $\tan 2 t = \frac{2 t}{1 - {t}^{2}}$ -->

$1 = \frac{2 t}{1 - {t}^{2}}$ --> ${t}^{2} + 2 t - 1 = 0$
$D = {d}^{2} = {b}^{2} - 4 a c = 4 + 4 = 8 - \to d = \pm 2 \sqrt{2}$

$t = \tan 112.5 = - \frac{2}{2} \pm \frac{2 \sqrt{2}}{2} = - 1 \pm \sqrt{2}$

Since t = 112.5 deg is in Quadrant II, its tan is negative, then only the negative answer is accepted : (-1 - sqrt2)