Prove a trigonometric equality

Aedrha2

New member
Joined
Jun 14, 2021
Messages
2
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!
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
24,988
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)
 
Last edited:

lex

Full Member
Joined
Mar 3, 2021
Messages
832
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]
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
24,988
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.
 
Top