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

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

#### Explanation:

We should start from the given

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

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

$\textcolor{red}{{\cos}^{2} \left(\frac{\theta}{4}\right) = \frac{1 + \cos \left(\frac{\theta}{2}\right)}{2} = \frac{1 + \sqrt{\frac{1 + \cos \theta}{2}}}{2}}$

For the second term using Half-Angle formulas for sine and cosine

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

$\cot \left(\frac{A}{2}\right) = \sqrt{\frac{1 + \cos A}{1 - \cos A}} = \sqrt{\left(\frac{1 + \cos A}{1 - \cos A}\right) \cdot \frac{\left(1 + \cos A\right)}{\left(1 + \cos A\right)}}$

$\cot \left(\frac{A}{2}\right) = \sqrt{{\left(1 + \cos A\right)}^{2} / \left(1 - {\cos}^{2} A\right)} = \sqrt{{\left(1 + \cos A\right)}^{2} / \left({\sin}^{2} A\right)}$

$\cot \left(\frac{A}{2}\right) = \frac{1 + \cos A}{\sin} A$

therefore

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

For the third term, using Half-Angle Formulas for Sine

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

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

therefore

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

$f \left(\theta\right) = {\cos}^{2} \left(\frac{\theta}{4}\right) + \cot \left(\frac{\theta}{2}\right) + {\sin}^{2} \left(\frac{\theta}{2}\right)$
$\textcolor{red}{f \left(\theta\right) = \frac{1 + \sqrt{\frac{1 + \cos \theta}{2}}}{2} + \frac{1 + \cos \theta}{\sin} \theta + \frac{1 - \cos \theta}{2}}$