# How do you simplify f(theta)=3cot(theta/2)+tan(theta/2)-3cos(theta/4) to trigonometric functions of a unit theta?

Aug 2, 2018

$2 \left(2 \csc \theta + \cot \theta\right) - \frac{3}{8} \sqrt{8 \sqrt{8}} \left(\sqrt{\left(2 \sqrt{2} \pm \sqrt{1 + \cos \theta}\right)}\right)$,
choosing $-$ of $\left(\pm\right)$, for $0 < \theta \in {Q}_{4}$.

#### Explanation:

$\frac{\theta}{4} \in {Q}_{1} \mathmr{and} \frac{\theta}{2} \in {Q}_{2}$,

for $0 < \theta \in {Q}_{4}$, and then, cos (theta/2) < 0.

f(theta) = 3cos (theta/2)/sin(theta/2) + sin(theta/2)/cos(theta/2)-3 ( sqrt (1/2(1 + cos(theta/2))

$= \frac{3 {\cos}^{2} \left(\frac{\theta}{2}\right) + {\sin}^{2} \left(\frac{\theta}{2}\right)}{\sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}$

-3 sqrt (1/2(1 + cos(theta/2))

$= \frac{3 \left(\frac{1}{2} \left(1 + \cos \theta\right)\right) + \frac{1}{2} \left(1 - \cos \theta\right)}{\frac{1}{2} \sin \theta}$
-3 (sqrt (1/2(1 +- sqrt(1/2(1+ cos theta ))))#

$= 2 \left(2 \csc \theta + \cot \theta\right) - \left(\frac{3}{8}\right) \sqrt{8 \sqrt{8}} \left(\sqrt{\left(2 \sqrt{2} \pm \sqrt{1 + \cos \theta}\right)}\right)$,

choosing $-$ of $\left(\pm\right)$, for $0 < \theta \in {Q}_{4}$.

Note that,

if $0 < \theta \in {Q}_{4} \in \left(- \frac{\pi}{2} , 0\right) , \frac{\theta}{2} \in {Q}_{4} \in \left(- \frac{\pi}{4} , 0\right)$

and, the choice is + ( for $\cos \left(\frac{\theta}{2}\right)$). Indeed, incredible.

I am not approving ( like my invalidating $r < 0$ ), use of negative

$\theta$. This contradiction is the reason.

As these are not in tune with the universal faith on conventions, I

do not invite any discussion. I would like to be alone as non-

conventional, on such foundation matters.