# Given costheta=24/25 and 270<theta<360, how do you find cos(theta/2)?

Apr 22, 2018

$\cos \left(\frac{\theta}{2}\right) = - \frac{7 \sqrt{2}}{10}$

#### Explanation:

The double angle formula is

$\cos 2 x = 2 {\cos}^{2} x - 1$

Solving for $\cos x$ yields the half angle formula,

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

So we know

$\cos \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1}{2} \left(\cos \theta + 1\right)}$$= \pm \sqrt{\frac{1}{2} \left(\frac{24}{25} + 1\right)} = \pm \sqrt{\frac{49}{50}}$

The question is slightly ambiguous on this point, but we're obviously talking about $\theta$ a positive angle in the fourth quadrant, meaning its half angle between ${135}^{\circ}$ and ${180}^{\circ}$ is in the second quadrant, so has a negative cosine.

We could be talking about the "same" angle but say it's between $- {90}^{\circ}$ and ${0}^{\circ}$ and then the half angle would be in the fourth quadrant with a positive cosine. That's why there's a $\pm$ in the formula.

In this problem we conclude

$\cos \left(\frac{\theta}{2}\right) = - \sqrt{\frac{49}{50}}$

That's a radical we can simplify a bit, let's say

$\cos \left(\frac{\theta}{2}\right) = - \sqrt{\frac{2 \left(49\right)}{100}} = - \frac{7}{10} \sqrt{2}$