nastiest

  1. M

    Challenge: The nastiest trig. identity that you have ever met!

    Prove the identities \frac{cos(n\theta) sin[\frac{(n+1)\theta}{2}]}{sin(\theta/2)} = \frac{cos[(n+1)\theta] cos\theta - cos\theta - cos[(n+1)\theta] + 1 + sin\theta sin[(n+1)\theta]}{2-2cos\theta} and \frac{sin(n\theta) - sin[(n+1)\theta] + sin\theta}{2-2cos\theta} =...