# How do you find sin(x/2), cos(x/2), and tan(x/2) from the given Cot(x) = 13?

May 11, 2018

#### Answer:

There are actually four values for $\frac{x}{2}$ on the unit circle, so four values for each trig function. The principal value of the half angle is around ${2.2}^{\circ} .$

$\cos \left(\frac{1}{2} \textrm{A r c} \textrm{\cot} 13\right) = \cos {2.2}^{\circ} = \sqrt{\frac{1}{2} \left(1 + \frac{13}{\sqrt{170}}\right)}$

$\sin \left(\frac{1}{2} \textrm{A r c} \textrm{\cot} 13\right) = \sin {2.2}^{\circ} = \sqrt{\frac{1}{2} \left(1 - \frac{13}{\sqrt{170}}\right)}$

$\tan \left(\frac{1}{2} \textrm{A r c} \textrm{\cot} 13\right) = \tan {2.2}^{\circ} = \sqrt{170} - 13$

Please see the explanation for the others.

#### Explanation:

Let's talk about the answer a bit first. There are two angles on the unit circle whose cotangent is $13$. One is around ${4.4}^{\circ}$, and another is that plus ${180}^{\circ}$, call it ${184.4}^{\circ}$. Each of those have two half angles, again separated by ${180}^{\circ} .$ The first one has half angles ${2.2}^{\circ}$ and ${182.2}^{\circ}$, the second has half angles ${92.2}^{\circ}$ and ${272.2}^{\circ}$, So there are really four half angles in question, with differing but related values for their trig functions.

We'll use the above angles as approximations so we have names for them.

Angles with cotangent of 13:

$\textrm{A r c} \textrm{\cot} 13 \approx {4.4}^{\circ}$

${180}^{\circ} + \textrm{A r c} \textrm{\cot} 13 \approx {184.4}^{\circ}$

Half angles:

$\frac{1}{2} \textrm{A r c} \textrm{\cot} 13 \approx {2.2}^{\circ}$

$\frac{1}{2} \left({360}^{\circ} + \textrm{A r c} \textrm{\cot} 13\right) \approx {182.2}^{\circ}$

$\frac{1}{2} \left({180}^{\circ} + \textrm{A r c} \textrm{\cot} 13\right) \approx {92.2}^{\circ}$

$\frac{1}{2} \left({360}^{\circ} + {180}^{\circ} + \textrm{A r c} \textrm{\cot} 13\right) \approx {272.2}^{\circ}$

OK, the double angle formulas for cosine are:

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

so the relevant half angle formulas are

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

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

That's all preliminary. Let's do the problem.

We'll do the tiny angle first, ${2.2}^{\circ} .$ We see the rest of them are just multiples of ${90}^{\circ}$ above that, so we can get their trig functions from this first angle.

A cotangent of 13 is a slope of $\frac{1}{13}$ so corresponds to a right triangle with opposite $1$, adjacent $13$ and hypotenuse $\sqrt{{13}^{2} + {1}^{2}} = \sqrt{170} .$

$\cos \left(\textrm{A r c} \textrm{\cot} 13\right) = \cos {4.4}^{\circ} = \frac{13}{\sqrt{170}}$

$\sin \left(\textrm{A r c} \textrm{\cot} 13\right) = \sin {4.4}^{\circ} = \frac{1}{\sqrt{170}}$

Now we apply the half angle formulas. For our teeny angle in the first quadrant, we choose the positive signs.

$\cos \left(\frac{1}{2} \textrm{A r c} \textrm{\cot} 13\right) = \cos {2.2}^{\circ} = \setminus \sqrt{\frac{1}{2} \left(1 + \cos \left({4.4}^{\circ}\right)\right)} = \sqrt{\frac{1}{2} \left(1 + \frac{13}{\sqrt{170}}\right)}$

We could try to simplify and move the fractions outside the radical, but I'm just going to leave it here.

$\sin \left(\frac{1}{2} \textrm{A r c} \textrm{\cot} 13\right) = \sin {2.2}^{\circ} = \setminus \sqrt{\frac{1}{2} \left(1 - \cos \left({4.4}^{\circ}\right)\right)} = \sqrt{\frac{1}{2} \left(1 - \frac{13}{\sqrt{170}}\right)}$

The tangent half angle is the quotient of those, but it's easier to use

$\tan \left(\frac{\theta}{2}\right) = \frac{\sin \theta}{1 + \cos \theta}$

$\tan \left(\frac{1}{2} \textrm{A r c} \textrm{\cot} 13\right) = \tan {2.2}^{\circ} = \frac{\frac{1}{\sqrt{170}}}{1 + \frac{13}{\sqrt{170}}} = \sqrt{170} - 13$

OK, that's all the hard part, but let's not forget the other angles.

$\cos {182.2}^{\circ} = - \cos {2.2}^{\circ} = - \sqrt{\frac{1}{2} \left(1 + \frac{13}{\sqrt{170}}\right)}$

$\sin {182.2}^{\circ} = - \sin {2.2}^{\circ} = - \sqrt{\frac{1}{2} \left(1 - \frac{13}{\sqrt{170}}\right)}$

$\tan {182.2}^{\circ} = \tan {2.2}^{\circ} = \sqrt{170} - 13$

Now we have the remaining angles, which swap sine and cosine, flipping signs. We won't repeat the forms except for tangent.

$\cos {92.2}^{\circ} = - \sin {2.2}^{\circ}$

$\sin {92.2}^{\circ} = \cos {2.2}^{\circ}$

$\tan {92.2}^{\circ} = - \frac{1}{\tan {2.2}^{\circ}} = - 13 - \sqrt{170}$

$\cos {272.2}^{\circ} = \sin {2.2}^{\circ}$

$\sin {272.2}^{\circ} = - \cos {2.2}^{\circ}$

$\tan {272.2}^{\circ} = \tan {92.2}^{\circ} = - 13 - \sqrt{170}$

Phew.

May 11, 2018

#### Answer:

color(indigo)(tan (x/2) = 0.0384, sin (x/2) = +-0.0384, cos (x/2) = +- 1

color(crimson)(tan (x/2) = -26.0384, sin (x/2) = +- 0.9993, cos (x/2) = +- 0.0384

#### Explanation:

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

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

+cos 2x = (1- 2tan^2 x) / (1 + tan^2 x)

$\cot x = \frac{1}{\tan} x = 13$

$\tan x = \frac{1}{13}$

tan x = 1 / 13 = (2 tan (x/2)) / (1 - tan^2 (x/2)#

$1 - {\tan}^{2} \left(\frac{x}{2}\right) = 26 \tan \left(\frac{x}{2}\right)$

$\tan \cdot 2 \left(\frac{x}{2}\right) + 26 \tan \left(\frac{x}{2}\right) - 1 = 0$

$\tan \left(\frac{x}{2}\right) = \frac{- 26 \pm \sqrt{{26}^{2} + 4}}{2}$

$\tan \left(\frac{x}{2}\right) = \frac{- 26 \pm \sqrt{680}}{2}$

$\tan \left(\frac{x}{2}\right) = 0.0384 , - 26.0384$

${\csc}^{2} x = 1 + {\cot}^{2} x$

$\therefore {\csc}^{2} \left(\frac{x}{2}\right) = 1 + {\cot}^{2} \left(\frac{x}{2}\right)$

But wen know $\cot \left(\frac{x}{2}\right) = \frac{1}{\tan} \left(\frac{x}{2}\right)$

When $\tan \left(\frac{x}{2}\right) = 0.0384$,

${\csc}^{2} \left(\frac{x}{2}\right) = 1 + {\left(\frac{1}{0.0384}\right)}^{2} = 679.1684$

$\csc \left(\frac{x}{2}\right) = \sqrt{679.1684} = \pm 26.0609$

$\sin \left(\frac{x}{2}\right) = \pm \left(\frac{1}{26.0609}\right) = \pm 0.0384$

$\cos \left(\frac{x}{2}\right) = \sin \frac{\frac{x}{2}}{\tan} \left(\frac{x}{2}\right) = \pm \frac{0.0384}{0.0384} = \pm 1$

When $\tan \left(\frac{x}{2}\right) = - 26.0384$,

${\csc}^{2} \left(\frac{x}{2}\right) = 1 + \left(\frac{1}{- 26.0384} ^ 2\right) = 1.0015$

$\sin \left(\frac{x}{2}\right) = \frac{1}{\sqrt{1.0015}} = \pm 0.9993$

$\cos \left(\frac{x}{2}\right) = \sin \frac{\frac{x}{2}}{\tan} \left(\frac{x}{2}\right) = \pm \frac{0.9993}{-} 26.0384 = \pm 0.0384$