化简 e^{pi(b-in)}-e^{-pi(b-in)}

解答

化简

e^{π (b - in)} - e^{- π (b - in)}

(- 1)^{n} e^{πb} - (- 1)^{n} e^{- πb}

求解步骤

e^{π (b - in)} - e^{- π (b - in)}

e^{π (b - in)} = (- 1)^{n} e^{πb}

e^{- π (b - in)} = (- 1)^{n} e^{- πb}

= (- 1)^{n} e^{πb} - (- 1)^{n} e^{- πb}

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