## Trigonometry (11th Edition) Clone

$$\cos195^\circ=-\frac{\sqrt{2+\sqrt3}}{2}$$
$$\cos195^\circ$$ We would write $195^\circ$ as the sum of $180^\circ$ and $15^\circ$, as the angle $15^\circ$ can be applied the half-angle identities. $$\cos195^\circ=\cos(180^\circ+15^\circ)$$ - Sum Identity for cosine: $\cos(A+B)=\cos A\cos B-\sin A\sin B$ with $A=180^\circ$ and $B=15^\circ$ $$\cos195^\circ=\cos180^\circ\cos15^\circ-\sin180^\circ\sin15^\circ$$ $\cos180^\circ=-1$ and $\sin180^\circ=0$ $$\cos195^\circ=-1\times\cos15^\circ-0\times\sin15^\circ$$ $$\cos195^\circ=-\cos15^\circ$$ *Calculate $\cos15^\circ$ $$\cos15^\circ=\cos\Big(\frac{30^\circ}{2}\Big)$$ Apply the identity $\cos\Big(\frac{A}{2}\Big)=\pm\sqrt{\frac{1+\cos A}{2}}$ with $A=30^\circ$. $$\cos15^\circ=\pm\sqrt{\frac{1+\cos30^\circ}{2}}$$ $$\cos15^\circ=\pm\sqrt{\frac{1+\frac{\sqrt3}{2}}{2}}$$ $$\cos15^\circ=\pm\sqrt{\frac{\frac{2+\sqrt3}{2}}{2}}$$ $$\cos15^\circ=\pm\sqrt{\frac{2+\sqrt3}{4}}$$ $$\cos15^\circ=\frac{\pm\sqrt{2+\sqrt3}}{2}$$ As $15^\circ$ is in quadrant I, where $\cos\theta\gt0$, so $\cos15^\circ\gt0$, so we would select the positive square root. $$\cos15^\circ=\frac{\sqrt{2+\sqrt3}}{2}$$ Therefore, $$\cos195^\circ=-\cos15^\circ=-\frac{\sqrt{2+\sqrt3}}{2}$$