Answer
$$\cos67.5^\circ=\frac{\sqrt{2-\sqrt2}}{2}$$
10 goes with B.
Work Step by Step
$$\cos67.5^\circ$$
$67.5^\circ$ is the half angle of $135^\circ$
$$67.5^\circ=\frac{135^\circ}{2}$$
Therefore, $$\cos67.5^\circ=\cos\Big(\frac{135^\circ}{2}\Big)$$
- Recall the half-angle identities:
$$\cos\Big(\frac{A}{2}\Big)=\pm\sqrt{\frac{1+\cos A}{2}}$$
Apply the identity to $\cos67.5^\circ$ with $A=135^\circ$, we have
$$\cos67.5^\circ=\pm\sqrt{\frac{1+\cos135^\circ}{2}}$$
The sign of $\cos67.5^\circ$ would determine whether positive or negative square root should be selected.
$67.5^\circ$ is in quadrant I, where $\cos\theta\gt0$. Thus, $\cos67.5^\circ\gt0$. We need to pick the positive square root as a result.
$$\cos67.5^\circ=\sqrt{\frac{1+\cos135^\circ}{2}}$$
As $135^\circ+45^\circ=180^\circ$, $|\cos135^\circ|=\cos45^\circ$
$135^\circ$ lies in quadrant II, where $\cos\theta\lt0$, meaning $\cos135^\circ\lt0$
Therefore, $\cos135^\circ=-\cos45^\circ=-\frac{\sqrt2}{2}$
So, $$\cos67.5^\circ=\sqrt{\frac{1-\frac{\sqrt2}{2}}{2}}$$
$$\cos67.5^\circ=\sqrt{\frac{\frac{2-\sqrt2}{2}}{2}}$$
$$\cos67.5^\circ=\sqrt{\frac{2-\sqrt2}{4}}$$
$$\cos67.5^\circ=\frac{\sqrt{2-\sqrt2}}{2}$$
10 goes with B.
