Mostre que \operatorname{sinh} z \pm \operatorname{sinh} w=2 \operatorname{sinh} \left(\frac{z \pm w}{2}\right) \operatorname{cosh} \left(\frac{z \mp w}{2}\right)
Passo 1
O enunciado pede para mostrarmos as fórmulas de soma e diferença para a função hiperbólica \operatorname{sinh}. Vamos precisar das definições via exponenciais:
\operatorname{sinh} x = \frac{e^{x} - e^{-x}}{2},
\operatorname{cosh} x = \frac{e^{x} + e^{-x}}{2}.
Passo 2
Derivando a soma \operatorname{sinh} z + \operatorname{sinh} w:
\operatorname{sinh} z + \operatorname{sinh} w = \frac{e^{z}-e^{-z}}{2} \;+\;\frac{e^{w}-e^{-w}}{2} = \frac{\,e^{z}+e^{w}-e^{-z}-e^{-w}\,}{2}.
\operatorname{sinh} z + \operatorname{sinh} w = 2\,\operatorname{sinh}\!\bigl(\tfrac{z+w}{2}\bigr)\operatorname{cosh}\!\bigl(\tfrac{z-w}{2}\bigr).
Passo 3
Derivando a diferença \operatorname{sinh} z - \operatorname{sinh} w:
\operatorname{sinh} z - \operatorname{sinh} w = \frac{e^{z}-e^{-z}}{2} \;-\;\frac{e^{w}-e^{-w}}{2} = \frac{\,e^{z}-e^{w}-\bigl(e^{-z}-e^{-w}\bigr)\,}{2}.
\operatorname{sinh} z - \operatorname{sinh} w = 2\,\operatorname{sinh}\!\bigl(\tfrac{z-w}{2}\bigr)\operatorname{cosh}\!\bigl(\tfrac{z+w}{2}\bigr).
Resposta
Portanto fica demonstrado que
\operatorname{sinh} z \pm \operatorname{sinh} w = 2\,\operatorname{sinh}\!\Bigl(\tfrac{z\pm w}{2}\Bigr)\operatorname{cosh}\!\Bigl(\tfrac{z\mp w}{2}\Bigr).