Since [tex]e^{ix}=\cos x+i\sin x[/tex], you have
[tex]\dfrac{(\sin x-i\cos x)^3}{(\cos x+i\sin x)^5}=\dfrac{(-ie^{ix})^3}{(e^{ix})^5}=\dfrac{ie^{3ix}}{e^{5ix}}=ie^{-2ix}[/tex]
So, [tex]|ie^{-2ix}|=|e^{-2ix}|=1[/tex] and the argument is [tex]-2x[/tex].
