Why is There a Problem with the Proof for sin(x/2) = +/- sqrt((1-cos(x))/2)?

[Uniquebum]

My problem is:
Proof $sin(\frac{x}{2}) = \pm \sqrt{\frac{1-cos(x)}{2}}$

Simple issue really i'd think but i can't come up with a way.

For starters i'd use however
$cos^2(x) + sin^2(x) = 1$ identity.

Which evidently would lead into
$sin(\frac{x}{2}) = \pm \sqrt{1-cos^2(\frac{x}{2})}$

But then i got nothing...

 

[Mentallic]

Do you know the expansion for cos(2x) in terms of sin(x) and cos(x)? From there you would convert this expression solely into terms with sin(x), and finally solve for sin(x).

 

[Uniquebum]

Ahhh i get it!
$cos(2\frac{x}{2}) = cos^2(\frac{x}{2})-sin^2(\frac{x}{2})$
$cos^2(\frac{x}{2}) = cos(2\frac{x}{2})-sin^2(\frac{x}{2})$

Thus
$sin^2(\frac{x}{2}) = 1-cos^2(\frac{x}{2})$
$sin^2(\frac{x}{2}) = 1-cos(x)-sin^2(\frac{x}{2})$

And so
$sin(\frac{x}{2}) = \pm \sqrt{\frac{1-cos(x)}{2}}$
Thanks!

 

[Mark44]

Ahhh i get it!

I'm not sure you do.

$cos(2\frac{x}{2}) = cos^2(\frac{x}{2})-sin^2(\frac{x}{2})$
$cos^2(\frac{x}{2}) = cos(2\frac{x}{2}) - sin^2(\frac{x}{2})$

No,
$cos^2(\frac{x}{2}) = cos(2\frac{x}{2}) + sin^2(\frac{x}{2})$

Thus
$sin^2(\frac{x}{2}) = 1-cos^2(\frac{x}{2})$
$sin^2(\frac{x}{2}) = 1-cos(x)-sin^2(\frac{x}{2})$

And so
$sin(\frac{x}{2}) = \pm \sqrt{\frac{1-cos(x)}{2}}$
Thanks!

 

[Uniquebum]

That was a typo but anyway... :)

 

[Mentallic]

Ahhh i get it!
$cos(2\frac{x}{2}) = cos^2(\frac{x}{2})-sin^2(\frac{x}{2})$
$cos^2(\frac{x}{2}) = cos(2\frac{x}{2})-sin^2(\frac{x}{2})$

Thus
$sin^2(\frac{x}{2}) = 1-cos^2(\frac{x}{2})$
$sin^2(\frac{x}{2}) = 1-cos(x)-sin^2(\frac{x}{2})$

And so
$sin(\frac{x}{2}) = \pm \sqrt{\frac{1-cos(x)}{2}}$
Thanks!

Nice work