From $\frac{1-\cos x}{\sin x}$ to $\tan\frac{x}{2}$ How can I write $\frac{1-\cos x}{\sin x}$ as $\tan\frac{x}{2}$? I wrote $\sin x$ as $2\sin\frac{x}{2} \cos\frac{x}{2}$ also used the double angle identity for $\cos$ but wasn't able to make much progress
 A: Also, $\cos x = \cos^2 (x/2) - \sin^2 (x/2)$.
Then letting $y=x/2$,
$$\frac{1-\cos x}{\sin x} = \frac{1 - \cos^2 y + \sin^2 y}{2 \sin y \cos y} = \frac{2 \sin^2 y}{2 \sin y \cos y} = \frac{\sin y}{\cos y} = \tan (x/2).$$
A: $$\cos(a+b)=\cos a \cos b -sin a sin b\\cos(x+x)=cos x \cos x -\sin x \sin x\\cos(2x)=cos^2x-sin^2x\\cos(2x)=cos^2x-(1-cos^2x)=2cos^2x-1\\or =(1-\sin^2x)-\sin^2x\\so\\1-cos(2x)=2sin^2x\\$$put there x instead of 2x $$1-cos(x)=2sin^2(\frac{x}{2})\\\frac{1-cos(x)}{sinx}=\frac{2sin^2(\frac{x}{2})}{2sin(\frac{x}{2})cos(\frac{x}{2})}=\frac{sin(\frac{x}{2})}{cos(\frac{x}{2})}=\\tan(\frac{x}{2})$$
A: Replacing 1 with $\cos\frac2{x}{2}$ + $\sin\frac{x}{2}$, $\cos$$x$ with  $\cos\frac2{x}{2}- \sin\frac{x}{2}$ and $\sin$$x$ with $2\sin\frac{x}{2}$$\cos\frac{x}{2}$ the numerator and denominator simplify to $2\sin^2\frac{x}{2}$ and $2\cos\frac{x}{2}\sin\frac{x}{2}$ and hence the fraction simplifies to $\tan\frac{x}{2}$.
A: you can do $\dfrac{1-\cos x}{\sin x} = \dfrac{\cos 0 - \cos x}{\sin 0 + \sin x}=\dfrac{2\cos x/2 \cos x/2}{2 \sin x/2 \cos x/2} = \tan(x/2).$  
we used $\cos a -\cos b = 2\cos (a-b)/2 \cos(a+b)/2$ and $\sin a + \sin b = 2 \sin(a+b)/2 \cos (a-b)/2$ with $a = 0$ and $b = x.$
