Answer
\[\sin 4x=\frac{i}{2}e^{-4ix}-\frac{i}{2}e^{4ix}\]
Work Step by Step
According to question we will use the result
\[\sin bx=\frac{1}{2i}\left(e^{ibx}-e^{-ibx}\right)\;\;\;\;\;\;\;....(1)\]
Using equation (1) with $b=4$
\[\sin 4x=\frac{1}{2i}\left(e^{i4x}-e^{-i4x}\right)\]
\[\sin 4x=\frac{1}{2i}e^{i4x}-\frac{1}{2i}e^{-i4x}\]
\[\sin 4x=i\frac{1}{2}e^{-i4x}-i\frac{1}{2}e^{i4x}\]
Hence $\;\;\;\sin 4x=\frac{i}{2}e^{-4ix}-\frac{i}{2}e^{4ix}$.
