I like Mathwonk's answer! A similar way to see that e^{ix}= cos(x)+ isin(x) is to use the Taylor's series at x= 0 (as jgens said):
e^{x}= 1+ x+ x^2/2+ x^3/6+ \cdot\cdot\cdot+ x^n/n!+ \cdot\cdot\cdot
cos(x)= 1- x^2/2+ x^4/4!+ \cdot\cdot\cdot+ (-1)^n x^{2n}/(2n!)+ \cdot\cdot\cdot
sin(x)= x- x^3/3!+ x^5/5!+ \cdot\cdot\cdot+ (-1)^n x^{2n+1}/(2n+1)!+ \cdot\cdot\cdot
Each involves powers of x over the factorial of that power. Sine has only odd powers, cosine only even powers and sine and cosine have alternating sign.
Now see what happens when we replace "x" with "ix" in the series for e^x:
e^{ix}= 1+ (ix)+ (ix)^2/2+ (ix)^3/3!+ \cdot\cdot\cdot+ (ix)^n/n!+ \cdot\cdot\cdot
e^{ix}= 1+ ix+ i^2x^2/2+ i^3x^3/3!+ \cdot\cdot\cdot+ i^nx^n/n!+ \cdot\cdot\cdot
Of course, i^2= -1 so i^3= i^2(x)= -i, i^4= -i(i)= 1 and then it all repeats. That is, every odd power of ix is plus or minus ix while every even power is plus or minus 1. So
e^{ix}= 1+ ix- x^2/2- ix^3/3!+ \cdot\cdot\cdot+ (-1)^nx^{2n}/(2n)!+ (-1)^ni x^{2n+1}/(2n+1)!+ \cdot\cdot\cdot
and separating "real" and "imaginary" parts
e^{ix}= (1- x^2/2+ x^4/4!+ \cdot\cdot\cdot+ (-1)^n x^{2n}/(2n)!+ \cdot\cdot\cdot)+ i(x- x^3/3!+ \cdot\cdot\cdot+ (-1)^n x^{2n+1}/(2n+1)!)
e^{ix}= cos(x)+ isin(x)
I don't know how to understand the exp term as sinusoidal? in my mind they simply go to zero or infinity.
That's because you are used to dealing with real numbers. The exponential is increasing along along the real axis, periodic along the imaginary axis.
