Any “shortcuts” to proving that $\frac{\sin(x)}2+\sin^2(\frac x2)\tan(\frac x2)\to\tan(\frac x2)$

Question

I was working on simplifying some trig functions, and after a while of playing with them I simplified $$\frac{\sin(x)}{2}+\sin^2\left(\frac{x}{2}\right)\tan\left(\frac{x}{2}\right) \rightarrow \tan\left(\frac{x}{2}\right)$$

The way I got that result, however, was with what I think a very "roundabout" way. I first used the half-angle formulaes, then used $x=\pi/2-\beta$, and that simplified to $$\frac{\cos(\beta)}{1+\sin(\beta)}$$ where I again used the coordinate change to get $$\frac{\sin(x)}{1+\cos(x)}\rightarrow\tan\left(\frac{x}{2}\right)$$

I tried using the online trig simplifiers but none succeeded. Of course, after you know the above identity, it's easy to prove by proving that $$\frac{\sin(x)}{2}=\tan\left(\frac{x}{2}\right)-\sin^2\left(\frac{x}{2}\right)\tan\left(\frac{x}{2}\right)$$

Is there a more direct way to get the identity? I guess what I'm asking is, am I missing any "tricks" or software that I could have on my toolbelt so that next time I don't spend hours trying to simplify trig identities?

Answer 1

With $s:=\sin\frac x2,c:=\cos\frac x2$, $\sin x=2sc$ and

$$\frac122sc+s^2\frac sc=\frac{sc^2+s^3}c=\frac sc.$$

Answer 2

Use $ \sin(x) = 2 \sin(x/2) \cos(x/2)$ then we have \begin{eqnarray*} \frac{\sin(x)}{2}+\sin^2\left(\frac{x}{2}\right)\tan\left(\frac{x}{2}\right) &=& \frac{\sin(x/2)}{\cos(x/2)} \underbrace{\left( \cos^2(x/2) + \sin^2(x/2) \right)}_{=1} \\ &=& \tan (x/2). \end{eqnarray*}

Answer 3

Fairly obvious: $$\sin^2\left(\frac{x}{2}\right)\tan\left(\frac{x}{2}\right)=(1-\cos^2\left(\frac{x}{2}\right))\tan\left(\frac{x}{2}\right)=\tan\left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)\sin\left(\frac{x}{2}\right)=\tan\left(\frac{x}{2}\right)-\frac{\sin x}{2}$$

Answer 4

$$ \frac{\sin x}{2} = \sin\frac{x}{2}\cos\frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} \cos^2\frac{x}{2} = \tan\frac{x}{2} \left(1 - \sin^2\frac{x}{2}\right) $$