How do you simplify f(theta)=2tan(theta/2)-sin(theta/4)-sec(theta/4) to trigonometric functions of a unit theta?

Apr 25, 2018

 f(theta) = pm 2 sqrt{ {1 - cos theta }/ { 1 + cos theta } } pm sqrt{1/2( 1pm sqrt{ 1 / 2 (1 - cos theta ) } ) } - 1/{pm sqrt{1/2( 1+ cos(theta/2)) } }

Explanation:

Hmm, let's derive the half angle formulas, which come from the cosine double angle formulas:

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

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

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

Let's set $x = \frac{\theta}{2}$

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

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

Let's set $x = \frac{\theta}{4}$

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

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

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

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

 f(theta) = { pm 2 sqrt{ 1 / 2 (1 - cos theta ) } }/ { pm sqrt{ 1 / 2 (1 + cos theta ) } } pm sqrt{1/2( 1pm sqrt{ 1 / 2 (1 - cos theta ) } ) } - 1/{pm sqrt{1/2( 1+ cos(theta/2)) } }

 f(theta) = pm 2 sqrt{ {1 - cos theta }/ { 1 + cos theta } } pm sqrt{1/2( 1pm sqrt{ 1 / 2 (1 - cos theta ) } ) } - 1/{pm sqrt{1/2( 1+ cos(theta/2)) } }

That doesn't seem particularly simplified to me, but there it is.