# Prove a trigonometric equality

#### Aedrha2

##### New member
Hey there!
I am trying to prove that: $\frac{\sin(2x)}{\sin(\frac{x}{2})}=2(\cos(\frac{3x}{2}+\cos(\frac{x}{2}))$

I can't seem to figure it out first i tried : $\frac{\sin(2x)}{\sin(\frac{x}{2})}=\frac{\sin(\frac{x}{2}+\frac{3x}{2})}{\sin(\frac{x}{2})}= \frac{\sin(\frac{x}{2})\cos(\frac{3x}{2})+\cos(\frac{x}{2})\sin(\frac{3x}{2})}{\sin(\frac{x}{2})}\\ \frac{\sin(\frac{x}{2})\cos(\frac{3x}{2})}{\sin(\frac{x}{2})}+\frac{\cos(\frac{x}{2})\sin(\frac{3x}{2})}{\sin(\frac{x}{2})}=\cos(\frac{3x}{2})+\frac{\cos(\frac{x}{2})\sin(\frac{3x}{2})}{\sin(\frac{x}{2})}$
Where I am missing a factor 2 and can't figure out what to do with the second term.

Then i tired:

$\frac{\sin(2x)}{\sin(\frac{x}{2})}=\frac{2\sin(x)\cos(x)}{\sin(\frac{x}{2})}$
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
Hey there!
I am trying to prove that: $\frac{\sin(2x)}{\sin(\frac{x}{2})}=2(\cos(\frac{3x}{2}+\cos(\frac{x}{2}))$

I can't seem to figure it out first i tried : $\frac{\sin(2x)}{\sin(\frac{x}{2})}=\frac{\sin(\frac{x}{2}+\frac{3x}{2})}{\sin(\frac{x}{2})}= \frac{\sin(\frac{x}{2})\cos(\frac{3x}{2})+\cos(\frac{x}{2})\sin(\frac{3x}{2})}{\sin(\frac{x}{2})}\\ \frac{\sin(\frac{x}{2})\cos(\frac{3x}{2})}{\sin(\frac{x}{2})}+\frac{\cos(\frac{x}{2})\sin(\frac{3x}{2})}{\sin(\frac{x}{2})}=\cos(\frac{3x}{2})+\frac{\cos(\frac{x}{2})\sin(\frac{3x}{2})}{\sin(\frac{x}{2})}$
Where I am missing a factor 2 and can't figure out what to do with the second term.

Then i tired:

$\frac{\sin(2x)}{\sin(\frac{x}{2})}=\frac{2\sin(x)\cos(x)}{\sin(\frac{x}{2})}$
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:
• Aedrha2, topsquark and lex

#### lex

##### Full Member
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]

• Aedrha2 and Subhotosh Khan

#### Subhotosh Khan

##### Super Moderator
Staff member
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.