MathCastle
TrigonometryChallengeThe Unit Circle and Circular Functions

The Unit Circle and Circular Functions · Challenge problem

Evaluate exactly: ∑k=111sin⁡2kπ12=sin⁡2π12+sin⁡22π12+⋯+sin⁡211π12\displaystyle\sum_{k=1}^{11}\sin^2\dfrac{k\pi}{12}=\sin^2\dfrac{\pi}{12}+\sin^2\dfrac{2\pi}{12}+\cdots+\sin^2\dfrac{11\pi}{12}.