To find $\cos(165^\circ)$, let's use the cosine function's properties:

1. Convert $165^\circ$ to an angle in a reference form: Since $165^\circ$ is in the second quadrant, we know that $\cos(165^\circ)$ will be negative, and we can use the reference angle $180^\circ - 165^\circ = 15^\circ$.

2. Cosine Value: Now we calculate $\cos(165^\circ) = -\cos(15^\circ)$.

3. Using the cosine of 15°: We know that $\cos(15^\circ) = \cos(45^\circ - 30^\circ) = \cos(45^\circ) \cos(30^\circ) + \sin(45^\circ) \sin(30^\circ)$ Using known values: $\cos(45^\circ) = \frac{\sqrt{2}}{2}, \quad \cos(30^\circ) = \frac{\sqrt{3}}{2}, \quad \sin(45^\circ) = \frac{\sqrt{2}}{2}, \quad \sin(30^\circ) = \frac{1}{2}$

4. Calculate $\cos(15^\circ)$: $\cos(15^\circ) = \left(\frac{\sqrt{2}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2}\right) \left(\frac{1}{2}\right)$ Simplifying, $= \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$

5. Final Answer: Therefore, $\cos(165^\circ) = -\frac{\sqrt{6} + \sqrt{2}}{4}$

So, $\cos(165^\circ) = -\frac{\sqrt{6} + \sqrt{2}}{4}$.

Would you like further details or have any questions?

Here are some related questions:

1. How do you find the sine of an angle in the second quadrant?
2. How do reference angles help in trigonometric calculations?
3. What is the difference between radians and degrees?
4. How does the cosine function behave in different quadrants?
5. How can trigonometric identities simplify calculations?

Tip: Remember, angles in the second quadrant always have a negative cosine value!