Proving $\frac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} = \left(\tan\frac{x}{2}\right)^2$ 
How do I prove this equality?
$$\frac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} = \left(\tan\frac{x}{2}\right)^2$$

I have come this far by myself:
$$\begin{array}{llll} 
\dfrac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} &= \dfrac{2\sin x- 2\sin x\cos x}{2\sin x+ 2\sin x\cos x} & \text{since $\sin(2x) = 2\sin x\cos x$}&\\
& = \dfrac{2\sin x(1 - \cos x)}{2\sin x(1 + \cos x)} &&\\
& = \dfrac{(1- \cos x)}{(1+ \cos x)} &&\\
& = \dfrac{(1- \cos x)(1+ \cos x)}{(1+ \cos x)(1+ \cos x)}& \text{since    $\dfrac{(1+ \cos x)}{(1+ \cos x)}=1$}&\\
& = \dfrac{(1)^2-(\cos x)^2}{(1+ \cos x)^2} & \text{since $a^2-b^2 = (a+b)(a-b)$}&\\
& = \dfrac{(\sin x)^2}{(1+ \cos x)^2} & \text{since $(\sin x)^2 + (\cos x)^2 =1$, so $(\sin x)^2 = 1- (\cos x)^2$.}&
\end{array}$$
Now, I understand that I have the $\sin x$ part on the numerator. What I have to do is get the denominator to be $\cos x$ somehow and also make the angles $\frac{x}{2}$ instead of $x$. How do I do that?
Please be through, and you can't use half-angle or triple angle or any of those formulas.
Also, we have to show left hand side is equal to right hand side, we can't do it the other way around. So please do not take $(\tan\frac{x}{2})^2$ and solve the equation.
Thank you for understanding and have a nice day :)
 A: You have proved that the LHS is 
$$
LHS=\frac{1-\cos x}{1+\cos x}
$$
And if you know the formula for $\tan \frac{x}{2}$ you have done (bisection).
If not use:
$$
\tan^2 \frac{x}{2}=\frac{\sin ^2 \frac{x}{2}}{\cos ^2 \frac{x}{2}}=\frac{\frac{1-\cos x}{2}}{\frac{1+\cos x}{2}}
$$
A: All you've done is fine.
Now: since we want $\frac{x}{2}$ angles to appear, use $\sin x=2\sin\frac{x}{2}\cos\frac{x}{2}$ and $1+\cos x=\cos 0+\cos x=2(\cos\frac{x}{2})^2$ (a formula for the sum of cosines), and you're done.


you cant use half-angle or triple angle or any of those formulas.

I don't understand - how to obtain $\tan\frac{x}{2}$ without using half-angles? ... since $\frac{x}{2}$ is in the answer, it has to appear somehow..?
A: Hint
If we cannot use the half angle, let us use the double angle formulae letting $x=2y$ $$\frac{1-\cos(x)}{1+\cos(x)}=\frac{1-\cos(2y)}{1+\cos(2y)}$$ Simplify and when finished reset $y=\frac x2$.
A: We have used here $$\sin { 2x } =2\sin { x } \cos { x } \\ \cos { x } =\cos ^{ 2 }{ \frac { x }{ 2 }  } -\sin ^{ 2 }{ \frac { x }{ 2 }  } \\ \cos ^{ 2 }{ \frac { x }{ 2 }  } +\sin ^{ 2 }{ \frac { x }{ 2 }  } =1\\ $$$$\frac { 2\sin { x } -\sin { 2x }  }{ 2\sin { x+\sin { 2x }  }  } =\frac { 2\sin { x } \left( 1-\cos { x }  \right)  }{ 2\sin { x } \left( 1+\cos { x }  \right)  } =\frac { \cos ^{ 2 }{ \frac { x }{ 2 } +\sin ^{ 2 }{ \frac { x }{ 2 } -\cos ^{ 2 }{ \frac { x }{ 2 } +\sin ^{ 2 }{ \frac { x }{ 2 }  }  }  }  }  }{ \cos ^{ 2 }{ \frac { x }{ 2 } +\sin ^{ 2 }{ \frac { x }{ 2 } +\cos ^{ 2 }{ \frac { x }{ 2 } -\sin ^{ 2 }{ \frac { x }{ 2 }  }  }  }  }  } =\\ =\frac { 2\sin ^{ 2 }{ \frac { x }{ 2 }  }  }{ 2\cos ^{ 2 }{ \frac { x }{ 2 }  }  } =\tan ^{ 2 }{ \frac { x }{ 2 }  } $$
A: You know that  $1-\cos x=2\sin^2\dfrac{x}2$   and  $1+\cos x=2\cos^2\dfrac{x}{2}$, so $$\frac{1-\cos x}{1+\cos x}=\frac{2}{2}\left(\frac{\sin\dfrac{x}2}{\cos\dfrac{x}2}\right)^2=\tan^2\frac{x}2.$$
If you couldn't understand, please comment.
A: Please see the attached photo for a solution.

