Mostre que
\operatorname{cosh} z-\operatorname{cosh} w=2 \operatorname{sinh} \left(\frac{z+w}{2}\right) \operatorname{sinh} \left(\frac{z-w}{2}\right)

Passo 1

Vamos provar a identidade
\operatorname{cosh} z - \operatorname{cosh} w = 2 \operatorname{sinh}\!\Bigl(\tfrac{z+w}{2}\Bigr)\,\operatorname{sinh}\!\Bigl(\tfrac{z-w}{2}\Bigr).
O conteúdo que vamos usar é a definição das funções hiperbólicas em termos de exponenciais:
\operatorname{cosh} u = \frac{e^u + e^{-u}}{2} \quad\text{e}\quad \operatorname{sinh} u = \frac{e^u - e^{-u}}{2}.

Passo 2

Escrevendo a diferença usando exponenciais, temos
\operatorname{cosh} z - \operatorname{cosh} w = \frac{e^z + e^{-z}}{2} \;-\; \frac{e^w + e^{-w}}{2} = \frac{e^z - e^w + e^{-z} - e^{-w}}{2}.

Passo 3

Separe as duas diferenças exponenciais e expresse-as em termos de \operatorname{sinh}:
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} = 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).

Passo 4

Somando essas duas expressões e dividindo por 2, obtemos
\operatorname{cosh} z - \operatorname{cosh} w = \frac{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)\,\bigl(2\,\operatorname{sinh}\!\bigl(\tfrac{z+w}{2}\bigr)\bigr),
ou seja,
\operatorname{cosh} z - \operatorname{cosh} w = 2 \operatorname{sinh}\!\Bigl(\tfrac{z+w}{2}\Bigr)\,\operatorname{sinh}\!\Bigl(\tfrac{z-w}{2}\Bigr).

Resposta

\operatorname{cosh} z - \operatorname{cosh} w = 2 \operatorname{sinh}\!\Bigl(\tfrac{z+w}{2}\Bigr)\,\operatorname{sinh}\!\Bigl(\tfrac{z-w}{2}\Bigr).

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