# Conversion of a trigonometic function

1. Mar 24, 2012

### frensel

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

2. Relevant equations

3. 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
$\tan(x)\sin(\frac{x}{2})+\cos(\frac{x}{2}) = \frac{\cos(\frac{x}{2})}{\cos(x)}$
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 24, 2012

### tiny-tim

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

3. Mar 24, 2012

### frensel

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?

4. Mar 24, 2012

### tiny-tim

you could work backwards (from the answer) …

(tan / 2sin1/2) - cos1/2 = … ?