# Chapter 5 - Trigonometric Identities - Section 5.6 Half-Angle Identities - 5.6 Exercises - Page 242: 15

$$\cos165^\circ=-\frac{\sqrt{2+\sqrt 3}}{2}$$

#### Work Step by Step

$$A=\cos165^\circ$$ We can separate $165^\circ$ into the difference of $180^\circ$ and $15^\circ$. Therefore, we can rewrite A as follows $$A=\cos(180^\circ-15^\circ)$$ $$A=\cos180^\circ\cos15^\circ+\sin180^\circ\sin15^\circ$$ $$A=-1\cos15^\circ+0\sin15^\circ$$ $$A=-\cos15^\circ$$ Since $15^\circ$ is half of $30^\circ$, we can rewrite $$A=-\cos\Bigg(\frac{1}{2}30^\circ\Bigg)$$ $$A=-\Bigg[\pm\sqrt{\frac{1+\cos30^\circ}{2}}\Bigg]$$ However, we know that $15^\circ$ is in the first quadrant, which means $\cos15^\circ\gt0$, so $$A=-\Bigg[\sqrt{\frac{1+\cos30^\circ}{2}}\Bigg]$$ $$A=-\sqrt\frac{1+\frac{\sqrt 3}{2}}{2}$$ $$A=-\sqrt\frac{\frac{2+\sqrt 3}{2}}{2}$$ $$A=-\sqrt\frac{2+\sqrt 3}{4}$$ $$A=-\frac{\sqrt{2+\sqrt 3}}{2}$$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.