Answer
\[\cos 8x=\frac{1}{2}e^{8ix}+\frac{1}{2}e^{-8ix}\]
Work Step by Step
According to question we will use the result
\[\cos bx=\frac{1}{2}\left(e^{ibx}+e^{-ibx}\right)\;\;\;\;\;\;\;....(1)\]
Using equation (1) with $b=8$
\[\cos 8x=\frac{1}{2}\left(e^{i8x}+e^{-i8x}\right)\]
\[\cos 8x=\frac{1}{2}e^{8ix}+\frac{1}{2}e^{-8ix}\]
Hence $\;\cos 8x=\frac{1}{2}e^{8ix}+\frac{1}{2}e^{-8ix}$.
