Prove a trigonometric equality

[Aedrha2]

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!

 

[Deleted member 4993]

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!

First I think you have a missing parenthesis on the right hand side.

Did you try to simplify the right-hand-side? If I were to solve this problem I would try to do that!

cos (x+x/2) + cos(x-x/2) = 2* cos(x) * cos(x/2)

 

[lex]

Alternatively you could prove the equivalent [imath]\tfrac{1}{2} \sin 2x = \cos\left(\tfrac{3x}{2}\right)\sin\left(\tfrac{x}{2}\right) +\cos\left(\tfrac{x}{2}\right) \sin\left(\tfrac{x}{2}\right)[/imath]
working with the right hand side using the identity: [imath]\hspace1ex \cos \alpha \sin \beta \equiv \tfrac{1}{2}(\sin (\alpha+\beta) - \sin (\alpha-\beta))[/imath]

 

[Deleted member 4993]

Alternatively you could prove the equivalent [imath]\tfrac{1}{2} \sin 2x = \cos\left(\tfrac{3x}{2}\right)\sin\left(\tfrac{x}{2}\right) +\cos\left(\tfrac{x}{2}\right) \sin\left(\tfrac{x}{2}\right)[/imath]
working with the right hand side using the identity: [imath]\hspace1ex \cos \alpha \sin \beta \equiv \tfrac{1}{2}(\sin (\alpha+\beta) - \sin (\alpha-\beta))[/imath]

I was hoping the student would discover your first solution.

 

[Deleted member 4993]

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!

Did you get unstuck?

 

[Aedrha2]

Yes i did get unstuck. Something like this:

[math]\frac{\sin(2\omega)}{\sin(\frac{\omega}{2})}=\frac{\sin(2\omega)+\sin(\omega)-\sin(\omega)+0}{\sin(\frac{\omega}{2})}\\ \frac{\sin(2\omega)+\sin(\omega)-\sin(\omega)+\sin0}{\sin(\frac{\omega}{2})}\\ \frac{\sin(\frac{\omega}{2}+\frac{3\omega}{2})-\sin(\frac{3\omega}{2}-\frac{\omega}{2})+\sin(\frac{\omega}{2}+\frac{\omega}{2})+\sin(\frac{\omega}{2}-\frac{\omega}{2})}{\sin(\frac{\omega}{2})}\\ \frac{2\cos(\frac{3\omega}{2})\sin(\frac{\omega}{2})+2\cos(\frac{\omega}{2})\sin(\frac{\omega}{2})}{\sin(\frac{\omega}{2})}\\ 2(\cos(\frac{3\omega}{2})+\cos(\frac{\omega}{2}))[/math]