# trigonometric transformation $\frac{(\cos x+i\sin x )-1}{(\cos x+i\sin x )+1}=-i\tan (x/2)$

I found this trigonometric part in a complex variable problem, can anyone please show how to get it,thanks

$$\frac{(\cos x+i\sin x )-1}{(\cos x+i\sin x )+1}=-i\tan (x/2)$$

• When $x=\frac \pi 2$, this would lead to $\frac{1-y}{1+y}=-y$ which does not show real roots. Post the original problem, please. – Claude Leibovici Aug 8 '17 at 7:55
• here y was not a variable actually it's i – tharanga dharmapala Aug 8 '17 at 7:58
• – Blue Aug 8 '17 at 7:59
• The first expression was true ,$$\frac{(\cos x+i\sin x )-1}{(\cos x+i\sin x )+1}=-i\tan (x/2)$$ if you put $x=\frac{\pi}{2}$ you will find $i-1=-(i-1)$ ! so go back to the first one – Khosrotash Aug 8 '17 at 8:22

$$\dfrac{1-(\cos x+i\sin x)}{1+(\cos x+i\sin x)}=\\ \dfrac{1-\cos x-i\sin x}{1+\cos x+i\sin x}=\\\dfrac{2sin^2(\frac x2)-i2sin(\frac x2)\cos(\frac x2)}{2\cos^2(\frac x2)+i2\sin(\frac x2)cos(\frac x2)}=$$simplify $$=\frac{\sin(\frac x2)}{\cos(\frac x2)}.\frac{sin(\frac x2)-icos(\frac x2)}{\cos(\frac x2)+i\sin(\frac x2)}=\\\tan(\frac x2)\times \frac ii\times \frac{\sin(\frac x2)-i\cos(\frac x2)}{\cos(\frac x2)+i\sin(\frac x2)}=\\\frac 1i \tan(\frac x2)\frac{i \sin(\frac x2)- i^2\cos(\frac x2)}{\cos(\frac x2)+isin(\frac x2)} =\\\frac 1i \tan(\frac x2)\frac{i \sin(\frac x2)+\cos(\frac x2)}{\cos(\frac x2)+i\sin(\frac x2)} =\\\frac 1i \tan(\frac x2)=\\-i\tan(\frac x2)$$
$$\dfrac{1-(\cos(x)+i\sin(x))}{1+(\cos(x)+i\sin(x))}=\frac{1-e^{ix} }{1+e^{ix} }=-\frac{e^{ix}-1 }{e^{ix}+1}=-\frac{e^{ix/2}-e^{-ix/2}}{e^{ix/2}+e^{-ix/2}}=-\tanh(i\frac x2)$$