Exercise 1576Algebra › Complex numbersOlympiadAdded 3 Nov 2026 Sum of cosines #trigonometric sum#cosines#complex StatementCompute ∑k=0n−1cos(kθ)\displaystyle\sum_{k=0}^{n-1}\cos(k\theta)k=0∑n−1cos(kθ) for θ∈R\theta\in\mathbb{R}θ∈R. HintTake the real part of the geometric series ∑k=0n−1eikθ\sum_{k=0}^{n-1}e^{ik\theta}∑k=0n−1eikθ. Show solutionHide solutionSolutionComplex sum.∑k=0n−1eikθ=1−einθ1−eiθ=ei(n−1)θ/2sin(nθ/2)sin(θ/2).\sum_{k=0}^{n-1}e^{ik\theta}=\frac{1-e^{in\theta}}{1-e^{i\theta}}=e^{i(n-1)\theta/2}\frac{\sin(n\theta/2)}{\sin(\theta/2)}.k=0∑n−1eikθ=1−eiθ1−einθ=ei(n−1)θ/2sin(θ/2)sin(nθ/2).Real part.∑k=0n−1cos(kθ)=sin(nθ/2)sin(θ/2)cos((n−1)θ2).\sum_{k=0}^{n-1}\cos(k\theta)=\frac{\sin(n\theta/2)}{\sin(\theta/2)}\cos\left(\frac{(n-1)\theta}{2}\right).k=0∑n−1cos(kθ)=sin(θ/2)sin(nθ/2)cos(2(n−1)θ).Conclusion. ∑k=0n−1cos(kθ)=sin(nθ/2)sin(θ/2)cos(n−1)θ2\displaystyle\sum_{k=0}^{n-1}\cos(k\theta)=\frac{\sin(n\theta/2)}{\sin(\theta/2)}\cos\frac{(n-1)\theta}{2}k=0∑n−1cos(kθ)=sin(θ/2)sin(nθ/2)cos2(n−1)θ (for sin(θ/2)≠0\sin(\theta/2)\neq0sin(θ/2)=0).