## Trigonometry (11th Edition) Clone

$$\cos67.5^\circ=\frac{\sqrt{2-\sqrt2}}{2}$$ 10 goes with B.
$$\cos67.5^\circ$$ $67.5^\circ$ 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.