## Trigonometry (11th Edition) Clone

$$2\cos^267\frac{1^\circ}{2}-1=-\frac{\sqrt2}{2}$$
$$2\cos^267\frac{1^\circ}{2}-1$$ 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 $2\cos^267\frac{1^\circ}{2}-1$ $$2\cos^267\frac{1^\circ}{2}-1=\cos135^\circ=-\frac{\sqrt2}{2}$$