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
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?