Finding a Trig Identity to Show a Relation

[indigojoker]

I know that \frac{1-cos(x)}{2sin\left(\frac{x}{2}\right)} = sin\left(\frac{x}{2}\right)

but is there a trig identity that states this? I've been manipulating a certain equation to try and fit a trig identity to make everything make sense. Actually, I started out with:

\frac{1-cos(x)}{\sqrt{2-2cos(x)}}

and got that to equal \frac{1-cos(x)}{2sin\left(\frac{x}{2}\right)}

not sure if there's a trig identity to show this relation.

 

[dynamicsolo]

I'm not entirely sure what you're asking for, but I'll guess you might be looking for this.
Start from the "double-angle formula" for cosine:

cos 2t = [(cos t)^2] - [(sin t)^2] = 1 - 2[(sin t)^2] ;

rearrange it to obtain

2[(sin t)^2] = 1 - cos 2t

and replace 2t with x .

The one thing you want to be a bit careful with is that the "half-angle formulas" for sine and cosine have a sign ambiguity, so you will have quadrant-dependent cases to consider for your \frac{1-cos(x)}{2sin\left(\frac{x}{2}\right)} = sin\left(\frac{x}{2}\right) ; sometimes a minus sign will be needed.

But in passing from

\frac{1-cos(x)}{\sqrt{2-2cos(x)}} to

\frac{1-cos(x)}{2sin\left(\frac{x}{2}\right)} ,

shouldn't that "2" in the denominator be "square root of 2"? (There is still the sign ambiguity, of course...)

 

[roadrunner]

toaly off the topic but how do i post the white boxes with formula s in them?! :S

 

[Gib Z]

Well if you can't derive the identity, you can still prove it, but I'm not sure if you know differentiation yet. If you do, just show that their derivatives are the same, and the constant they differ by is 0.

However the result follows easily off the power reduction formulae!

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

That formula is in turn derived from the double angle expansion of cos 2x,\cos (2x) = 1- 2\sin^2 x.

 

[HallsofIvy]

toaly off the topic but how do i post the white boxes with formula s in them?! :S

You mean LaTex? There is a tutorial on "LaTex Math Typesetting" in the tutorials section. Go back from this forum to the Education forum and click on "Tutorials".

 

[indigojoker]

thanks dynamicsolo,

there's another trig identity problem that I'm trying to figure out...

\frac{sin(x)}{2sin(x/2)}=cos(x/2)

any ideas?

 

[dynamicsolo]

thanks dynamicsolo,

there's another trig identity problem that I'm trying to figure out...

\frac{sin(x)}{2sin(x/2)}=cos(x/2)

any ideas?

Multiply through by the denominator (with the usual proviso that there are certain values where it is zero). Does this expression look familiar? How about if you replace x with 2t?

 

[genneth]

\sin(2x) = 2 \cos x \sin x

Btw, you should use \sin instead of just tying sin in tex mode... My inner typesetter just rebels against incorrect typography. :blush: