1. **Problem Statement:** Find the value of the expression $$2 \cos \left(\frac{\pi}{13}\right) \cos \left(\frac{9\pi}{13}\right) + \cos \left(\frac{3\pi}{13}\right) + \cos \left(\frac{5\pi}{13}\right)$$. 2. **Formula and Important Rules:** We use the product-to-sum and sum-to-product identities for cosine: - $$\cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)]$$ - Sum of cosines can sometimes be simplified using known angle sum identities or symmetry properties. 3. **Step-by-step Solution:** - First, apply the product-to-sum formula to $$2 \cos \left(\frac{\pi}{13}\right) \cos \left(\frac{9\pi}{13}\right)$$: $$2 \cos \left(\frac{\pi}{13}\right) \cos \left(\frac{9\pi}{13}\right) = 2 \times \frac{1}{2} \left[ \cos \left(\frac{\pi}{13} + \frac{9\pi}{13}\right) + \cos \left(\frac{\pi}{13} - \frac{9\pi}{13}\right) \right] = \cos \left(\frac{10\pi}{13}\right) + \cos \left(-\frac{8\pi}{13}\right)$$ - Since $$\cos(-x) = \cos x$$, this becomes: $$\cos \left(\frac{10\pi}{13}\right) + \cos \left(\frac{8\pi}{13}\right)$$ - Now the entire expression is: $$\cos \left(\frac{10\pi}{13}\right) + \cos \left(\frac{8\pi}{13}\right) + \cos \left(\frac{3\pi}{13}\right) + \cos \left(\frac{5\pi}{13}\right)$$ - Group terms: $$\left[ \cos \left(\frac{3\pi}{13}\right) + \cos \left(\frac{10\pi}{13}\right) \right] + \left[ \cos \left(\frac{5\pi}{13}\right) + \cos \left(\frac{8\pi}{13}\right) \right]$$ - Use the sum formula for cosines: $$\cos A + \cos B = 2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$$ - For the first bracket: $$2 \cos \left(\frac{3\pi/13 + 10\pi/13}{2}\right) \cos \left(\frac{3\pi/13 - 10\pi/13}{2}\right) = 2 \cos \left(\frac{13\pi}{26}\right) \cos \left(-\frac{7\pi}{26}\right) = 2 \cos \left(\frac{\pi}{2}\right) \cos \left(\frac{7\pi}{26}\right)$$ - Since $$\cos \left(\frac{\pi}{2}\right) = 0$$, the first bracket sums to 0. - For the second bracket: $$2 \cos \left(\frac{5\pi/13 + 8\pi/13}{2}\right) \cos \left(\frac{5\pi/13 - 8\pi/13}{2}\right) = 2 \cos \left(\frac{13\pi}{26}\right) \cos \left(-\frac{3\pi}{26}\right) = 2 \cos \left(\frac{\pi}{2}\right) \cos \left(\frac{3\pi}{26}\right)$$ - Again, $$\cos \left(\frac{\pi}{2}\right) = 0$$, so the second bracket sums to 0. 4. **Final Answer:** Adding both brackets gives 0. **Therefore, the value of the expression is** $0$. **Answer choice:** (1) 0