Conversion of a trigonometic function

[frensel]

Homework Statement

How to convert
\tan(x)\sin(\frac{x}{2})+\cos(\frac{x}{2})
to
\frac{\tan(x)}{\sqrt{2(1-\cos(x))}}

Homework Equations

The Attempt at a Solution

I can convert it to this form: \frac{\cos(\frac{x}{2})}{\cos(x)}
\tan(x)\sin(\frac{x}{2})+\cos(\frac{x}{2})
=\frac{\sin(x)}{\cos(x)}\sin(\frac{x}{2})+ \cos(\frac{x}{2})
=\frac{1}{\cos(x)}\left(\sin(x)\sin(\frac{x}{2})+ \cos(x)\cos(\frac{x}{2})\right)
using angle sum and difference identities, we get
\left(\sin(x)\sin(\frac{x}{2})+ \cos(x)\cos(\frac{x}{2})\right) = \cos(x - \frac{x}{2}) = \cos(\frac{x}{2})
therefore, we have
\tan(x)\sin(\frac{x}{2})+\cos(\frac{x}{2}) = \frac{\cos(\frac{x}{2})}{\cos(x)}

 

[tiny-tim]

hi frensel

I can convert it to this form: \frac{\cos(\frac{x}{2})}{\cos(x)}

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

 

[frensel]

hi frensel

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

I got it, thx!

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

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

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

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

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 = … ?