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}.
Agora fatoramos e^{z}+e^{w} e e^{-z}+e^{-w}:
e^{z}+e^{w} = e^{\frac{z+w}{2}}\bigl(e^{\frac{z-w}{2}}+e^{-\frac{z-w}{2}}\bigr) = 2\,e^{\frac{z+w}{2}}\operatorname{cosh}\!\bigl(\tfrac{z-w}{2}\bigr),
e^{-z}+e^{-w} = e^{-\frac{z+w}{2}}\bigl(e^{\frac{z-w}{2}}+e^{-\frac{z-w}{2}}\bigr) = 2\,e^{-\frac{z+w}{2}}\operatorname{cosh}\!\bigl(\tfrac{z-w}{2}\bigr).
Então o numerador fica
2\,\operatorname{cosh}\!\bigl(\tfrac{z-w}{2}\bigr)\bigl(e^{\frac{z+w}{2}}-e^{-\frac{z+w}{2}}\bigr) = 2\,\operatorname{cosh}\!\bigl(\tfrac{z-w}{2}\bigr)\cdot 2\operatorname{sinh}\!\bigl(\tfrac{z+w}{2}\bigr),
e, dividindo por 2, obtemos
\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}.
Fatoramos agora
e^{z}-e^{w} = e^{\frac{z+w}{2}}\bigl(e^{\frac{z-w}{2}}-e^{-\frac{z-w}{2}}\bigr) = 2\,e^{\frac{z+w}{2}}\operatorname{sinh}\!\bigl(\tfrac{z-w}{2}\bigr),
e^{-z}-e^{-w} = -\bigl(e^{-\frac{z+w}{2}}\bigr)\bigl(e^{\frac{z-w}{2}}-e^{-\frac{z-w}{2}}\bigr) = -2\,e^{-\frac{z+w}{2}}\operatorname{sinh}\!\bigl(\tfrac{z-w}{2}\bigr).
Assim o numerador vira
2\,\operatorname{sinh}\!\bigl(\tfrac{z-w}{2}\bigr)\bigl(e^{\frac{z+w}{2}}+e^{-\frac{z+w}{2}}\bigr) = 2\,\operatorname{sinh}\!\bigl(\tfrac{z-w}{2}\bigr)\cdot 2\operatorname{cosh}\!\bigl(\tfrac{z+w}{2}\bigr),
e, dividindo por 2, chega-se a
\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).

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