TrigonometrySuperchallengeThe Unit Circle and Circular Functions ShareThe Unit Circle and Circular Functions · Superchallenge problemLet h(t)=cost1+2th(t) = \dfrac{\cos t}{1 + 2^t}h(t)=1+2tcost. Evaluate exactly:∑k=−66h (kπ6)=h(−π)+h (−5π6)+⋯+h (5π6)+h(π).\sum_{k=-6}^{6} h\!\left(\frac{k\pi}{6}\right) = h(-\pi) + h\!\left(-\frac{5\pi}{6}\right) + \cdots + h\!\left(\frac{5\pi}{6}\right) + h(\pi).k=−6∑6h(6kπ)=h(−π)+h(−65π)+⋯+h(65π)+h(π).Check Show answer Walk through it step by step Practice more like this Review this concept