TrigonometryChallengeThe Unit Circle and Circular Functions ShareThe Unit Circle and Circular Functions · Challenge problemEvaluate exactly: ∑k=111sin2kπ12=sin2π12+sin22π12+⋯+sin211π12\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}k=1∑11sin212kπ=sin212π+sin2122π+⋯+sin21211π.Check Show answer Walk through it step by step Practice more like this Review this concept