# How can you solve for tan theta in tan (theta) + tan 3 (theta) /1 -tan (theta) *3 (theta) = -1 with your answer in a surd form?

May 6, 2018

$\theta = - {11.25}^{\circ} + {45}^{\circ} k$

$\tan \theta =$

$1 + \sqrt{2} - \sqrt{2 \left(2 + \sqrt{2}\right)} \quad \mathmr{and}$

$1 - \sqrt{2} + \sqrt{4 - 2 \sqrt{2}} \quad \quad \quad \mathmr{and}$

$1 + \sqrt{2} + \sqrt{2 \left(2 + \sqrt{2}\right)} \quad \mathmr{and}$

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

#### Explanation:

I'm going to guess this is supposed to read: Solve for $\tan \theta$ in

$\frac{\tan \theta + \tan 3 \theta}{1 - \tan \theta \tan 3 \theta} = - 1$

We recognize this as the tangent sum angle formula:

$\tan \left(a + b\right) = \frac{\tan a + \tan b}{1 - \tan a \tan b}$

So our equation becomes,

$\tan 4 \theta = - 1$

It's tempting to try to jump right to the answer, but a prefer a standard approach that uses the fact that $\cos x = \cos a$ has solutions $x = \setminus \pm a + {360}^{\circ} k \quad$ for integer $k$.

$\frac{\sin 4 \theta}{\cos 4 \theta} = - 1$

$- \sin 4 \theta = \cos 4 \theta$

$\sin \left(- 4 \theta\right) = \cos 4 \theta$

$\cos \left({90}^{\circ} + 4 \theta\right) = \cos 4 \theta$

${90}^{\circ} + 4 \theta = \setminus \pm 4 \theta + {360}^{\circ} k$

Only the minus sign gets us solutions:

$8 \theta = - {90}^{\circ} + {360}^{\circ} k$

$\theta = - {11.25}^{\circ} + {45}^{\circ} k$

EDIT: Oops, I solved for $\theta$ when I should have solved for $\tan \theta .$

DId you ever notice there's no half angle formula for tangent in terms of tangent? (There is one we can get with the quadratic formula.) The typical path is through the sine and cosine.

$\tan \left(\frac{a}{2}\right) = \frac{1 - \cos a}{\sin} a$

This part is tricky because there are two values for the pair $\left(\cos a , \sin a\right)$ that give any given tangent. In one the cosine and the sine are negated relative to the other.

$\cos \left(\frac{a}{2}\right) = \pm \sqrt{\frac{1}{2} \left(1 + \cos a\right)}$

sin(a/2) = pm sqrt{1/2(1-cos a)}

We have eight $\theta$s that will give rise to four unique values of $\tan \theta$. Let's calculate them.

$4 \theta = - {45}^{\circ} \mathmr{and} {145}^{\circ}$

$\cos \left(- {45}^{\circ}\right) = \frac{1}{\sqrt{2}} \quad \quad \quad \quad \sin \left(- {45}^{\circ}\right) = - \frac{1}{\sqrt{2}}$

$\cos \left(- {45}^{\circ} / 2\right) = \setminus \sqrt{\frac{1}{2} \left(1 + \frac{\sqrt{2}}{2}\right)}$

$\cos \left(- {22.5}^{\circ}\right) = \frac{1}{2} \sqrt{2 + \sqrt{2}}$

$\sin \left(- {22.5}^{\circ}\right) = - \frac{1}{2} \sqrt{2 - \sqrt{2}}$

$\tan \left(- {11.25}^{\circ}\right) = \frac{1 - \cos \left(- {22.5}^{\circ}\right)}{\sin \left(- {22.5}^{\circ}\right)} = \frac{1 - \frac{1}{2} \sqrt{2 + \sqrt{2}}}{- \frac{1}{2} \sqrt{2 - \sqrt{2}}}$$= 1 + \sqrt{2} - \sqrt{2 \left(2 + \sqrt{2}\right)}$

$\tan \left(a + {45}^{\circ}\right) = \frac{\tan a + \tan {45}^{\circ}}{1 - \tan a \tan {45}^{\circ}} = \frac{1 + \tan a}{1 - \tan a}$

 tan(-11.25^circ + 45^circ) = tan(33.75^circ) ={1 + ( 1 + sqrt(2) - sqrt(2 (2 + sqrt(2))) )}/{1 - ( 1 + sqrt(2) - sqrt(2 (2 + sqrt(2))) ) } = 1 - sqrt(2) + sqrt(4 - 2 sqrt(2))

$\tan \left(- {11.25}^{\circ} + {90}^{\circ}\right) = \frac{1}{\tan} \left(- {11.25}^{\circ} / 2\right) =$$= 1 + \sqrt{2} + \sqrt{2 \left(2 + \sqrt{2}\right)}$

$\tan \left(a - {45}^{\circ}\right) = \frac{\tan a - 1}{\tan a + 1}$

$\tan \left(- {11.25}^{\circ} - {45}^{\circ}\right) = \tan \left(- {56.25}^{\circ}\right) = \frac{\left(1 + \sqrt{2} - \sqrt{2 \left(2 + \sqrt{2}\right)}\right) - 1}{\left(1 + \sqrt{2} - \sqrt{2 \left(2 + \sqrt{2}\right)}\right) + 1}$ = 1 - sqrt(2) (1 + sqrt(2 - sqrt(2)))

May 6, 2018

$t = \frac{3 \pi}{16} + \frac{k \pi}{4}$, or
$t = {33}^{\circ} 75 + k {90}^{\circ}$

#### Explanation:

$\tan 4 t = \tan \left(t + 3 t\right) = \frac{\tan t + \tan 3 t}{1 - \tan t . \tan 3 t} = - 1$
tan 4t = -1
Trig table and unit circle give ->
$4 t = \frac{3 \pi}{4} + k \pi$
$t = \frac{3 \pi}{16} + \frac{k \pi}{4}$, or in degrees -->
$t = {33}^{\circ} 75 + k {90}^{\circ}$
Check by calculator.
t = 33.75 + 90 = 123.75 --> 4t = 495 --> tan 4t = - 1. Proved.