\( \frac{\text { Example }}{\text { Prove that } \cos 80^{\circ}+\cos 40^{\circ}=\cos 20^{\circ}} \)

**Step 1.** Recall the sum-to-product formula for cosine: \[ \cos A + \cos B = 2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right) \] **Step 2.** Let \( A = 80^\circ \) and \( B = 40^\circ \). Substitute these into the formula: \[ \cos 80^\circ + \cos 40^\circ = 2 \cos \left(\frac{80^\circ + 40^\circ}{2}\right) \cos \left(\frac{80^\circ - 40^\circ}{2}\right) \] **Step 3.** Simplify the expression: \[ \frac{80^\circ + 40^\circ}{2} = \frac{120^\circ}{2} = 60^\circ \] \[ \frac{80^\circ - 40^\circ}{2} = \frac{40^\circ}{2} = 20^\circ \] Thus, \[ \cos 80^\circ + \cos 40^\circ = 2 \cos 60^\circ \cos 20^\circ \] **Step 4.** Recall that \( \cos 60^\circ = \frac{1}{2} \). Substitute this value: \[ \cos 80^\circ + \cos 40^\circ = 2 \cdot \frac{1}{2} \cdot \cos 20^\circ = \cos 20^\circ \] **Conclusion.** Therefore, we have proven that: \[ \cos 80^\circ + \cos 40^\circ = \cos 20^\circ \]