Given that \( \cos (A+B)+\cos (A-B)=2 \cos A \cos B \). a) Prove the above identity. b) Hence, determine the value of \( \cos 80^{\circ}+\cos 40^{\circ} \) if \( \cos 20^{\circ}=0,94 \)

To prove the identity \( \cos (A+B) + \cos (A-B) = 2 \cos A \cos B \), we can use the cosine addition and subtraction formulas: 1. Start by applying the cosine addition formula: \[ \cos (A+B) = \cos A \cos B - \sin A \sin B \] 2. Now apply the cosine subtraction formula: \[ \cos (A-B) = \cos A \cos B + \sin A \sin B \] 3. Add these two results: \[ \cos (A+B) + \cos (A-B) = (\cos A \cos B - \sin A \sin B) + (\cos A \cos B + \sin A \sin B) \] 4. Simplifying this gives: \[ \cos (A+B) + \cos (A-B) = 2 \cos A \cos B \] Thus, the identity is proven. **Now to determine \( \cos 80^{\circ} + \cos 40^{\circ} \)**: Using the identity we've just proven, we set \( A = 60^{\circ} \) and \( B = 20^{\circ} \): \[ \cos(60^{\circ} + 20^{\circ}) + \cos(60^{\circ} - 20^{\circ}) = 2 \cos 60^{\circ} \cos 20^{\circ} \] \[ \cos(80^{\circ}) + \cos(40^{\circ}) = 2 \cdot \frac{1}{2} \cdot \cos 20^{\circ} = \cos 20^{\circ} \] Given that \( \cos 20^{\circ} = 0.94 \): \[ \cos 80^{\circ} + \cos 40^{\circ} = 0.94 \] So, the value of \( \cos 80^{\circ} + \cos 40^{\circ} \) is \( 0.94 \).