We need to evaluate: \[ \cos 20^\circ + \cos 100^\circ + \cos 140^\circ \] Step 1: Using the sum-to-product identity The sum-to-product identity for two cosines is: \[ \cos A + \cos B = 2 \cos \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \] We pair terms: \[ \cos 20^\circ + \cos 140^\circ \] Using the identity with \( A = 20^\circ \) and \( B = 140^\circ \): \[ \cos 20^\circ + \cos 140^\circ = 2 \cos \left( \frac{20^\circ + 140^\circ}{2} \right) \cos \left( \frac{20^\circ - 140^\circ}{2} \right) \] \[ = 2 \cos \left( \frac{160^\circ}{2} \right) \cos \left( \frac{-120^\circ}{2} \right) \] \[ = 2 \cos 80^\circ \cos (-60^\circ) \] Since \( \cos(-\theta) = \cos \theta \), we get: \[ = 2 \cos 80^\circ \cos 60^\circ \] \[ = 2 \cos 80^\circ \times \frac{1}{2} \] \[ = \cos 80^\circ \] Step 2: Adding \( \cos 100^\circ \) Now we compute: \[ \cos 80^\circ + \cos 100^\circ \] Using the sum-to-product identity with \( A = 80^\circ \) and \( B = 100^\circ \): \[ \cos 80^\circ + \cos 100^\circ = 2 \cos \left( \frac{80^\circ + 100^\circ}{2} \right) \cos \left( \frac{80^\circ - 100^\circ}{2} \right) \] \[ = 2 \cos 90^\circ \cos (-10^\circ) \] Since \( \cos 90^\circ = 0 \), the entire expression evaluates to 0.