conversion of a trigonometic function

frensel

1. The problem statement, all variables and given/known data
How to convert
[itex]\tan(x)\sin(\frac{x}{2})+\cos(\frac{x}{2})[/itex]
to
[itex]\frac{\tan(x)}{\sqrt{2(1-\cos(x))}}[/itex]


2. Relevant equations



3. The attempt at a solution
I can convert it to this form: [itex]\frac{\cos(\frac{x}{2})}{\cos(x)}[/itex]
[itex]\tan(x)\sin(\frac{x}{2})+\cos(\frac{x}{2})[/itex]
=[itex]\frac{\sin(x)}{\cos(x)}\sin(\frac{x}{2})+ \cos(\frac{x}{2})[/itex]
=[itex]\frac{1}{\cos(x)}\left(\sin(x)\sin(\frac{x}{2})+ \cos(x)\cos(\frac{x}{2})\right)[/itex]
using angle sum and difference identities, we get
[itex]\left(\sin(x)\sin(\frac{x}{2})+ \cos(x)\cos(\frac{x}{2})\right) = \cos(x - \frac{x}{2}) = \cos(\frac{x}{2})[/itex]
therefore, we have
[itex]\tan(x)\sin(\frac{x}{2})+\cos(\frac{x}{2}) = \frac{\cos(\frac{x}{2})}{\cos(x)}[/itex]
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

tiny-tim

hi frensel hint: multiply by sin(x)/sin(x), and use the half-angle identities

frensel

I got it, thx!

[itex]\frac{\cos(\frac{x}{2})}{\cos{x}}[/itex]
[itex] = \frac{\sin(x)}{\sin(x)}\frac{\cos(\frac{x}{2})}{ \cos{x}}[/itex]
[itex]=\tan(x)\frac{\cos(\frac{x}{2})}{\sin(x)}[/itex]

using double-angle formula, we have
[itex]\tan(x)\frac{\cos(\frac{x}{2})}{\sin(x)}[/itex]
[itex]=\tan(x)\frac{\cos(\frac{x}{2})}{2\sin(\frac{x}{2})\cos(\frac{x}{2})}[/itex]
[itex]=\tan(x)\frac{1}{2\sin(\frac{x}{2})}[/itex]

finally, using half-angle formula (assuming [itex]\sin(\frac{x}{2})>0[/itex]), then

[itex]\tan(x)\frac{1}{2\sin(\frac{x}{2})}[/itex]
[itex]=\tan(x)\frac{1}{2\sqrt{\frac{1-\cos(x)}{2}}}[/itex]
[itex]=\frac{\tan(x)}{\sqrt{2(1-\cos(x))}}[/itex]

Well, although I get the correct result, the calculation is so complicated. Is there any easier way to convert the above trigonometric function?

tiny-tim

you could work backwards (from the answer) …

(tan / 2sin1/2) - cos1/2 = … ?