MHB (e^ix)-1=(2ie^(ix/2))x(sin (x/2))

[minachi12]

Show that:
(e^ix)-1=(2ie^(ix/2))x(sin (x/2))

Calculate this sum
Zn=1+e^(ix)+e^(2ix)+...+e^((n-1)ix)
And deduce those values :
Xn=1+cos(x)+cos(2x)+...+cos((n-1)x)
Yn=sin(x)+sin(2x)+...+sin((n-1)x)

 

[HOI]

Show that:
(e^ix)-1=(2ie^(ix/2))x(sin (x/2))

Is that "x" in the middle a multiplication sign? Don't do that!

You need to know that $sin(x)= \frac{e^{ix}- e^{-ix}}{2i}$. So $sin(x/2)= \frac{e^{ix/2}- e^{-ix/2}}{2i}$, $sin(x/2)= \frac{e^{ix/2}- e^{-ix/2}}{2i}$, $2i e^{ix/2} sin(x/2)= e^{ix/2}(e^{ix/2}- e^{-ix/2})= e^{ix/2+ ix/2}- e^{ix/2- ix/2}= e^{ix}- 1$.

Calculate this sum
Zn=1+e^(ix)+e^(2ix)+...+e^((n-1)ix)
And deduce those values :
Xn=1+cos(x)+cos(2x)+...+cos((n-1)x)
Yn=sin(x)+sin(2x)+...+sin((n-1)x)

That's a geometric series, $\sum_{i= 0}^n r^i$ with $r= e^{ix}$. The sum of such a geometric series is $\frac{1- r^n}{1- r}$