Mostre que
\sin z \pm \sin w=2 \sin \left(\frac{z \pm w}{2}\right) \cos \left(\frac{z \mp w}{2}\right)
\begin{aligned} \sin z \pm \sin w & =\frac{e^{i z}-e^{-i z}}{2 i} \pm \frac{e^{i w}-e^{-i w}}{2 i}=\frac{e^{i \frac{z+w}{2}}\left(e^{i \frac{z-w}{2}} \pm e^{-i \frac{z-w}{2}}\right) \mp e^{-i \frac{z+w}{2}}\left(e^{i \frac{z-w}{2}} \pm e^{-i \frac{z-w}{2}}\right)}{2 i} \\\ & =2 \frac{\left(e^{i \frac{z+w}{2}} \mp e^{-i \frac{z+w}{2}}\right)\left(e^{i \frac{z-w}{2}} \pm e^{-i \frac{z-w}{2}}\right)}{2 \cdot 2 i}=2 \sin \left(\frac{z \pm w}{2}\right) \cos \left(\frac{z \mp w}{2}\right) . \end{aligned}

Passo 1

O enunciado pede para provar a identidade trigonométrica para números (ou variáveis complexas) z e w:
\sin z \pm \sin w = 2 \,\sin\!\Bigl(\frac{z \pm w}{2}\Bigr)\,\cos\!\Bigl(\frac{z \mp w}{2}\Bigr).
Vamos usar a representação de \sin em termos de exponenciais complexas.

Passo 2

Escrevemos as definições:
\sin z = \frac{e^{i z} - e^{-i z}}{2i}, \qquad \sin w = \frac{e^{i w} - e^{-i w}}{2i}.

Passo 3

Formamos a soma ou diferença:
\sin z \pm \sin w = \frac{e^{i z} - e^{-i z}}{2i} \pm \frac{e^{i w} - e^{-i w}}{2i} = \frac{1}{2i}\Bigl(e^{i z}\pm e^{i w} - (e^{-i z}\pm e^{-i w})\Bigr).

Passo 4

Agrupamos os termos exponenciais em fatores centrais. Observe que
e^{i z}\pm e^{i w} = e^{i\frac{z+w}{2}}\bigl(e^{i\frac{z-w}{2}}\pm e^{-i\frac{z-w}{2}}\bigr),
e^{-i z}\pm e^{-i w} = e^{-i\frac{z+w}{2}}\bigl(e^{-i\frac{z-w}{2}}\pm e^{i\frac{z-w}{2}}\bigr).
Substituindo em \sin z \pm \sin w, vem
\sin z \pm \sin w = \frac{1}{2i}\Bigl[ e^{i\frac{z+w}{2}}\bigl(e^{i\frac{z-w}{2}}\pm e^{-i\frac{z-w}{2}}\bigr) -e^{-i\frac{z+w}{2}}\bigl(e^{-i\frac{z-w}{2}}\pm e^{i\frac{z-w}{2}}\bigr) \Bigr].

Passo 5

Reconhecemos as fórmulas
e^{iA}-e^{-iA}=2i\,\sin A, \qquad e^{iB}+e^{-iB}=2\,\cos B.
No nosso caso, tomamos
A = \frac{z\pm w}{2}, \qquad B = \frac{z\mp w}{2}.
Então o numerador fica
2i\,\sin\!\Bigl(\tfrac{z\pm w}{2}\Bigr)\;\times\;2\,\cos\!\Bigl(\tfrac{z\mp w}{2}\Bigr) =4i\,\sin\!\Bigl(\tfrac{z\pm w}{2}\Bigr)\cos\!\Bigl(\tfrac{z\mp w}{2}\Bigr).
Dividindo por 2i resulta em
\sin z \pm \sin w =2\,\sin\!\Bigl(\tfrac{z\pm w}{2}\Bigr)\cos\!\Bigl(\tfrac{z\mp w}{2}\Bigr).

Resposta

\sin z \pm \sin w = 2\,\sin\!\Bigl(\tfrac{z \pm w}{2}\Bigr)\,\cos\!\Bigl(\tfrac{z \mp w}{2}\Bigr).

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