Answer
$$2\cos^267\frac{1^\circ}{2}-1=-\frac{\sqrt2}{2}$$
Work Step by Step
$$2\cos^267\frac{1^\circ}{2}-1$$
Recall the Double-Angle Identity for cosine:
$$2\cos^2A-1=\cos2A$$
Therefore, replace $A=67\frac{1^\circ}{2}$, we can apply the above identity to $2\cos^267\frac{1^\circ}{2}-1$.
$$2\cos^267\frac{1^\circ}{2}-1=\cos\Big(2\times67\frac{1^\circ}{2}\Big)$$
$$2\cos^267\frac{1^\circ}{2}-1=\cos135^\circ$$
As $135^\circ+45^\circ=180^\circ$, the absolute value of $\cos135^\circ$ would equal the (already positive) value of $\cos45^\circ$.
$$|\cos135^\circ|=\cos45^\circ$$
However, $135^\circ$ lies in quadrant II, where $\cos\theta\lt0$, so $\cos135^\circ\lt0$. Thus,
$$\cos135^\circ=-\cos45^\circ=-\frac{\sqrt2}{2}$$
Now bring the value of $\cos135^\circ$ back to the calculation of $2\cos^267\frac{1^\circ}{2}-1$
$$2\cos^267\frac{1^\circ}{2}-1=\cos135^\circ=-\frac{\sqrt2}{2}$$
