Answer:
[tex]\cos'(z) = -\sin(z)[/tex]
Step-by-step explanation:
According to the information given by the problem
[tex]\sin(z) = {\displaystyle \frac{e^{iz} - e^{-iz} }{2i} }[/tex]
[tex]\cos(z) = {\displaystyle \frac{e^{iz} + e^{-iz} }{2} }[/tex]
Now, if you compute the derivative of [tex]\cos[/tex] you get that
[tex]\cos'(z) = {\displaystyle \frac{ ie^{iz}-i e^{iz} }{2} } = {\displaystyle \frac{ i ( e^{iz}- e^{iz} )}{2} }\\\\= {\displaystyle \frac{ i ( e^{iz}- e^{iz} )}{2} } *\frac{i}{i} }\\\\= {\displaystyle - \frac{ e^{iz}- e^{iz} }{2i} } = -\sin(z)[/tex]
