# Prove that tan x = (sin 2x)/(1+cos 2x) . Hence find the value of tan 15° and tan 67.5° , giving your answer in term of surds in simplest form?

Aug 28, 2015

Prove $\tan x = \frac{\sin 2 x}{1 + \cos 2 x}$

#### Explanation:

Reminder: $1 + \cos 2 x = 2 {\cos}^{2} x$ and $\sin 2 x = 2 \sin x . \cos x$
$\frac{\sin 2 x}{1 + \cos 2 x} = \frac{2 \sin x . \cos x}{2 {\cos}^{2} x} = \sin \frac{x}{\cos} x = \tan x$

Call tan 15 = tan t
$\tan 2 t = \tan 30 = \frac{1}{\sqrt{3}}$
Apply the trig identity: $\tan 2 t = \frac{2 \tan t}{1 - {\tan}^{2} t}$
$\tan 2 t = \frac{1}{\sqrt{3}} = \frac{2 \tan t}{1 - {\tan}^{2} t}$ We get a quadratic equation:
$1 - {\tan}^{2} t = 2 \sqrt{3.} \tan t$
${\tan}^{2} t + 2 \sqrt{3.} \tan t - 1 = 0.$
$D = {d}^{2} = {b}^{2} - 4 a c = 12 + 4 = 16$ --> $d = \pm 4$
$\tan t = - 2 \frac{\sqrt{3}}{2} \pm \frac{4}{2} = - \sqrt{3} \pm 2$
Since t = 15 deg (Quadrant I), tan 15 > 0, then,
$\tan 15 = \tan t = - \sqrt{3} + 2$

Call tan 67.5 = tan t
tan 2t = tan 135 = tan (-45 + 180) = -tan 45 = - 1
$- 1 = \frac{2 \tan t}{1 - {\tan}^{2} t}$
${\tan}^{2} t - 2 \tan t - 1 = 0$
$D = {d}^{2} = 4 + 4 = 8$ --> $d = \pm 2 \sqrt{2}$
$\tan 67.5 = \tan t = \frac{2}{2} \pm \frac{2 \sqrt{2}}{2} = 1 \pm \sqrt{2}$
Since the arc 67.5 is located in Quadrant I, then, tan 67.5 > 0,
$\tan 67.5 = 1 + \sqrt{2}$
Check by calculator.
tan 15 = 0.27 ; (-sqrt3 + 2) = 0.27. OK
tan 67.5 = 2.414 ; (1 + sqrt2) = 1 + 1.414 = 2.414. OK