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Prove (a) cosh2(x) β sinh2(x) = 1 and (b) 1 β tanh 2(x) = sech 2(x). solution (a) cosh2(x) β sinh2(x) = ex + eβx 2 2 β 2 = e2x + 2 + eβ2x 4 β = 4 = . (b) we start with the identity proved in part (a): cosh2(x) β sinh2(x) = 1. if we divide both sides by cosh2(x), we get 1 β sinh2(x) cosh2(x) = 1 or 1 β tanh 2(x) = .
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What is the sum of the geometric series in which a1 = 7, r = 3, and an = 1,701? hint: cap s sub n equals start fraction a sub one left parenthesis one minus r to the power of n end power right parenthesis over one minus r end fraction comma r β 1, where a1 is the first term and r is the common ratio
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