# Prove this formula $1+\cos\theta+\cos2\theta+...+\cos n\theta=\frac{1}{2}+\frac{\sin(n+\frac{1}{2})\theta}{2\sin\frac{\theta}{2}}$ [duplicate]

This is homework but I’m really stuck. The question is to prove a fromula which states:

$$1+\cos\theta+\cos2\theta+...+\cos n\theta=\frac{1}{2}+\frac{\sin(n+\frac{1}{2})\theta}{2\sin\frac{\theta}{2}}$$

I want to solve it using complex numbers So I came to this (I missed Re in last one) Can you guys please help me finish this ?

• Multiply top and bottom with $(1-\cos\theta)+i\sin\theta$. Commented Oct 4, 2020 at 6:44
• Commented Oct 4, 2020 at 7:21
• @MartinR thank you for your consideration. It did <3 Commented Oct 4, 2020 at 7:47
• @rtybase Thank you for your consideration,I already found an answer.but thank you for your time. I checked it .It was also helpful. Commented Oct 4, 2020 at 8:51

• $2Sin(-\frac{\theta}{2})=-2Sin(\frac{\theta}{2})$.In your answer ,seventh line , what happened to the - ? Commented Oct 4, 2020 at 7:34
• Cause we know that $2i Sinx=e^{ix}-e^{-ix}$ Commented Oct 4, 2020 at 7:38