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!
<br /> cos(2\frac{x}{2}) = cos^2(\frac{x}{2})-sin^2(\frac{x}{2})<br />
<br /> cos^2(\frac{x}{2}) = cos(2\frac{x}{2})-sin^2(\frac{x}{2})<br />

Thus
<br /> sin^2(\frac{x}{2}) = 1-cos^2(\frac{x}{2})<br />
<br /> sin^2(\frac{x}{2}) = 1-cos(x)-sin^2(\frac{x}{2})<br />

And so
<br /> sin(\frac{x}{2}) = \pm \sqrt{\frac{1-cos(x)}{2}}<br />
Thanks!

 

[Mark44]

Ahhh i get it!

I'm not sure you do.

<br /> cos(2\frac{x}{2}) = cos^2(\frac{x}{2})-sin^2(\frac{x}{2})<br />
<br /> cos^2(\frac{x}{2}) = cos(2\frac{x}{2}) - sin^2(\frac{x}{2})<br />

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

Thus
<br /> sin^2(\frac{x}{2}) = 1-cos^2(\frac{x}{2})<br />
<br /> sin^2(\frac{x}{2}) = 1-cos(x)-sin^2(\frac{x}{2})<br />

And so
<br /> sin(\frac{x}{2}) = \pm \sqrt{\frac{1-cos(x)}{2}}<br />
Thanks!

 

[Uniquebum]

That was a typo but anyway... :)

 

[Mentallic]

Ahhh i get it!
<br /> cos(2\frac{x}{2}) = cos^2(\frac{x}{2})-sin^2(\frac{x}{2})<br />
<br /> cos^2(\frac{x}{2}) = cos(2\frac{x}{2})-sin^2(\frac{x}{2})<br />

Thus
<br /> sin^2(\frac{x}{2}) = 1-cos^2(\frac{x}{2})<br />
<br /> sin^2(\frac{x}{2}) = 1-cos(x)-sin^2(\frac{x}{2})<br />

And so
<br /> sin(\frac{x}{2}) = \pm \sqrt{\frac{1-cos(x)}{2}}<br />
Thanks!

Nice work