Hey there!
I am trying to prove that: [math]\frac{\sin(2x)}{\sin(\frac{x}{2})}=2(\cos(\frac{3x}{2}+\cos(\frac{x}{2}))[/math]
I can't seem to figure it out first i tried : [math]\frac{\sin(2x)}{\sin(\frac{x}{2})}=\frac{\sin(\frac{x}{2}+\frac{3x}{2})}{\sin(\frac{x}{2})}= \frac{\sin(\frac{x}{2})\cos(\frac{3x}{2})+\cos(\frac{x}{2})\sin(\frac{3x}{2})}{\sin(\frac{x}{2})}\\ \frac{\sin(\frac{x}{2})\cos(\frac{3x}{2})}{\sin(\frac{x}{2})}+\frac{\cos(\frac{x}{2})\sin(\frac{3x}{2})}{\sin(\frac{x}{2})}=\cos(\frac{3x}{2})+\frac{\cos(\frac{x}{2})\sin(\frac{3x}{2})}{\sin(\frac{x}{2})}[/math]
Where I am missing a factor 2 and can't figure out what to do with the second term.
Then i tired:
[math]\frac{\sin(2x)}{\sin(\frac{x}{2})}=\frac{2\sin(x)\cos(x)}{\sin(\frac{x}{2})}[/math]
Where i have the factor two but don't really know how to go on.
I'm really stuck and would appreciate any hints or help!
Thank you!
I am trying to prove that: [math]\frac{\sin(2x)}{\sin(\frac{x}{2})}=2(\cos(\frac{3x}{2}+\cos(\frac{x}{2}))[/math]
I can't seem to figure it out first i tried : [math]\frac{\sin(2x)}{\sin(\frac{x}{2})}=\frac{\sin(\frac{x}{2}+\frac{3x}{2})}{\sin(\frac{x}{2})}= \frac{\sin(\frac{x}{2})\cos(\frac{3x}{2})+\cos(\frac{x}{2})\sin(\frac{3x}{2})}{\sin(\frac{x}{2})}\\ \frac{\sin(\frac{x}{2})\cos(\frac{3x}{2})}{\sin(\frac{x}{2})}+\frac{\cos(\frac{x}{2})\sin(\frac{3x}{2})}{\sin(\frac{x}{2})}=\cos(\frac{3x}{2})+\frac{\cos(\frac{x}{2})\sin(\frac{3x}{2})}{\sin(\frac{x}{2})}[/math]
Where I am missing a factor 2 and can't figure out what to do with the second term.
Then i tired:
[math]\frac{\sin(2x)}{\sin(\frac{x}{2})}=\frac{2\sin(x)\cos(x)}{\sin(\frac{x}{2})}[/math]
Where i have the factor two but don't really know how to go on.
I'm really stuck and would appreciate any hints or help!
Thank you!