# How do you find cos67.5 using the half-angle identity?

##### 2 Answers
Jul 25, 2016

$\cos \left({67.5}^{\circ}\right) = + \frac{\sqrt{2 - \sqrt{2}}}{2}$

#### Explanation:

The Half-Angle Identity is $1 + \cos \theta = 2 {\cos}^{2} \left(\frac{\theta}{2}\right)$

Taking, $\theta = 2 \times {67.5}^{\circ} = {135}^{\circ} , 1 + \cos \left({135}^{\circ}\right) = 2 {\cos}^{2} \left({67.5}^{\circ}\right)$

$\therefore 2 {\cos}^{2} \left({67.5}^{\circ}\right) = 1 + \cos \left({180}^{\circ} - {45}^{\circ}\right) = 1 - \cos {45}^{\circ}$

$= 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{2 - \sqrt{2}}{2}$

$\therefore {\cos}^{2} \left({67.5}^{\circ}\right) = \frac{2 - \sqrt{2}}{4}$

$\therefore \cos \left({67.5}^{\circ}\right) = + \frac{\sqrt{2 - \sqrt{2}}}{2}$, $+ v e$ sign, as, 67.5^@ is in the first Quadrant#

Jul 25, 2016

I got $\frac{\sqrt{2 - \sqrt{2}}}{2}$ as well.

Since ${67.5}^{\circ}$ is half of ${135}^{\circ}$, let's work off of that. I remember it by starting from the ${\cos}^{2} \left(x\right)$ identity.

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

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

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

Since ${67.5}^{\circ}$ is in quadrant $\text{I}$, where $0 < x < {90}^{\circ}$, and $\cos \left(x\right)$ for $0 < x < {90}^{\circ}$ is positive, $\cos \left({67.5}^{\circ}\right) > 0$. Therefore:

$\cos \left({135}^{\circ} / 2\right) = + \sqrt{\frac{1 + \cos \left({135}^{\circ}\right)}{2}}$

$= \sqrt{\frac{1 + \cos \left(\frac{3 \pi}{4}\right)}{2}}$

$= \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}}$

$= \sqrt{\frac{\frac{2}{2} - \frac{\sqrt{2}}{2}}{2}} = \sqrt{\frac{\frac{2 - \sqrt{2}}{2}}{2}}$

$= \sqrt{\frac{2 - \sqrt{2}}{4}}$

$= \textcolor{b l u e}{\frac{\sqrt{2 - \sqrt{2}}}{2} \approx 0.3827}$