How do you express f(theta)=2sin(theta/2)-5cos(-theta/2-pi) in terms of trigonometric functions of a whole theta?

Aug 2, 2018

$2 \sqrt{\frac{1}{2} \left(1 - \cos \theta\right)} - 5 \left(\pm \sqrt{\frac{1}{2} \left(1 + \cos \theta\right)}\right)$,
choosing $\left(-\right)$ of $\left(\pm\right)$, when
theta in ( 2kpi - pi/2, 2kpi }, k = 1, 2, 3,

Explanation:

$f \left(\theta\right) = 2 \sin \left(\frac{1}{2} \theta\right) - 5 \cos \left(- \frac{1}{2} \theta - \pi\right)$

$= 2 \sin \left(\frac{1}{2} \theta\right) - 5 \cos \left(\frac{1}{2} \theta + \pi\right)$

$= 2 \sin \left(\frac{1}{2} \theta\right) + 5 \cos \left(\frac{1}{2} \theta\right)$

$= 2 \sqrt{\frac{1}{2} \left(1 - \cos \theta\right)} - 5 \left(\pm \sqrt{\frac{1}{2} \left(1 + \cos \theta\right)}\right)$,

choosing $\left(-\right)$ of $\left(\pm\right)$, when

theta in ( 2kpi - pi/2, 2kpi }, k = 1, 2, 3, ... .

I measure theta as positive, in both anticlockwise and clockwise

rotations.

if theta in ( - pi/2, 0), theta/2 in ( - pi/4, 0 ).

the ambiguity in sign of

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

does not arise at all.

Mathematical exactitude is needed, for disambiguation.

Alone, I am an advocate of strict adherence to'' $r \ge 0$ and so is

$\theta$''.