# How do you express f(theta)=sin(theta/4)+tan(theta/2)+cot(theta/2) in terms of trigonometric functions of a whole theta?

Nov 29, 2016

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

#### Explanation:

We use the following

$\cos \theta = 2 {\cos}^{2} \left(\frac{\theta}{2}\right) - 1$

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

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

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

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

$= \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}}$

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

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

Similarly,

cot(theta/2)=cos(theta/2)/(sin(theta/2)

$= \sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}}$

$= \sqrt{\frac{1 + \cos \theta}{1 - {\cos}^{2} \theta}}$

$= \frac{1 + \cos \theta}{\sin} \theta$

$\sin \left(\frac{\theta}{4}\right) = \sqrt{\frac{1 - \cos \left(\frac{\theta}{2}\right)}{2}}$

$= \sqrt{\frac{1 - \sqrt{\frac{1 + \cos \theta}{2}}}{2}}$

And finally,

$f \left(\theta\right) = \sin \left(\frac{\theta}{4}\right) + \tan \left(\frac{\theta}{2}\right) + \cot \left(\frac{\theta}{2}\right)$

$= \sqrt{\frac{1 - \sqrt{\frac{1 + \cos \theta}{2}}}{2}} + \frac{1 - \cos \theta}{\sin} \theta + \frac{1 + \cos \theta}{\sin} \theta$

$= \sqrt{\frac{1 - \sqrt{\frac{1 + \cos \theta}{2}}}{2}} + \frac{2}{\sin} \theta$