Mathematics, 23.06.2019 16:30 Maaarii
Evaluate \dfrac14c+3d 4 1 c+3d when c=6c=6 and d=7d=7.
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Prove (a) cosh2(x) − sinh2(x) = 1 and (b) 1 − tanh 2(x) = sech 2(x). solution (a) cosh2(x) − sinh2(x) = ex + e−x 2 2 − 2 = e2x + 2 + e−2x 4 − = 4 = . (b) we start with the identity proved in part (a): cosh2(x) − sinh2(x) = 1. if we divide both sides by cosh2(x), we get 1 − sinh2(x) cosh2(x) = 1 or 1 − tanh 2(x) = .
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Evaluate \dfrac14c+3d 4 1 c+3d when c=6c=6 and d=7d=7....
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