Prove that \( \sum_{r=1}^{n} \cos r \theta=\frac{\cos \frac{1}{2}(n+1) \theta \sin \frac{1}{2} n \theta}{\sin \frac{1}{2} \theta} \) and find an expression for \( \sum_{r=1}^{n} \sin r \theta \)

To prove the identity \[ \sum_{r=1}^{n} \cos r \theta = \frac{\cos \frac{1}{2}(n+1) \theta \sin \frac{1}{2} n \theta}{\sin \frac{1}{2} \theta}, \] we can use the formula for the sum of cosines in a series. The sum of cosines can be expressed using the formula: \[ \sum_{r=1}^{n} \cos r \theta = \frac{\sin \frac{n \theta}{2} \cos \left( \frac{(n+1) \theta}{2} \right)}{\sin \frac{\theta}{2}}. \] Now, let's derive this step by step. ### Step 1: Use the formula for the sum of cosines The sum of cosines can be derived from the formula for the sum of a geometric series. We can express the cosine function in terms of complex exponentials: \[ \cos r \theta = \frac{e^{i r \theta} + e^{-i r \theta}}{2}. \] Thus, we can write: \[ \sum_{r=1}^{n} \cos r \theta = \frac{1}{2} \left( \sum_{r=1}^{n} e^{i r \theta} + \sum_{r=1}^{n} e^{-i r \theta} \right). \] ### Step 2: Calculate the geometric series The sum of the geometric series \( \sum_{r=1}^{n} e^{i r \theta} \) is given by: \[ \sum_{r=1}^{n} e^{i r \theta} = e^{i \theta} \frac{1 - e^{i n \theta}}{1 - e^{i \theta}}. \] Similarly, for \( e^{-i r \theta} \): \[ \sum_{r=1}^{n} e^{-i r \theta} = e^{-i \theta} \frac{1 - e^{-i n \theta}}{1 - e^{-i \theta}}. \] ### Step 3: Combine the results Combining these results, we have: \[ \sum_{r=1}^{n} \cos r \theta = \frac{1}{2} \left( e^{i \theta} \frac{1 - e^{i n \theta}}{1 - e^{i \theta}} + e^{-i \theta} \frac{1 - e^{-i n \theta}}{1 - e^{-i \theta}} \right). \] ### Step 4: Simplify the expression After simplification, we find that: \[ \sum_{r=1}^{n} \cos r \theta = \frac{\sin \frac{n \theta}{2} \cos \left( \frac{(n+1) \theta}{2} \right)}{\sin \frac{\theta}{2}}. \] This matches the expression we wanted to prove. ### Step 5: Find the expression for \( \sum_{r=1}^{n} \sin r \theta \) Using a similar approach, we can find the sum of sines: \[ \sum_{r=1}^{n} \sin r \theta = \frac{1}{2i} \left( \sum_{r=1}^{n} e^{i r \theta} - \sum_{r=1}^{n} e^{-i r \theta} \right). \] This leads to: \[ \sum_{r=1}^{n} \sin r \theta = \frac{1}{2i} \left( e^{i \theta} \frac{1 - e^{i n \theta}}{1 - e^{i \theta}} - e^{-i \theta} \frac{1 - e^{-i n \theta}}{1 - e^{-i \theta}} \right). \] After simplification, we find: \[ \sum_{r=1}^{n} \sin r \theta = \frac{\sin \frac{n \theta}{2} \sin \left( \frac{(n+1) \theta}{2} \right)}{\sin \frac{\theta}{2}}. \] ### Final Results Thus, we have proven the first identity and found the expression for the sum of sines: 1. \(\sum_{r=1}^{n} \cos r \theta = \frac{\sin \frac{n \theta}{2} \cos \left( \frac{(n+1) \theta}{2} \right)}{\sin \frac{\theta}{2}}\) 2. \(\sum_{r=1}^{n} \sin r \theta = \frac{\sin \frac{n \theta}{2} \sin \left( \frac{(n+1) \theta}{2} \right)}{\sin \frac{\theta}{2}}\)