Which Sign do I choose using the Half Angle Formula for sin for this? I'm evaluating $\sin\left(\frac{1}{2}\sin^{-1}\left(-\frac{7}{25}\right)\right).$
The first thing I did was rewrite it as $\sin\left(\frac{\beta }{2}\right)$
Then I said that 
$\sin\left(\beta \right)=-\frac{7}{25}$
Using the Pythagorean Identity I found $cos\left(\beta \right)$
$\cos\left(\beta \right)=\pm\sqrt{1-\left(-\frac{7}{25}\right)^2}$
So $\cos\left(\beta \right)=\pm\frac{24}{25}$
Then to choose the sign of $\cos\left(\beta \right)$, I did this:
1) $\sin^{-1}\left(...\right)$: QI or QIV
2) $\sin\left(\beta \right)>0$   QI or QII
3) $\rightarrow \cos\left(\beta \right)$ is in QI
$\cos\left(\beta \right)=+\frac{24}{25}$
Then I applied this to the Half Angle Formula for Sine:
$\pm\sqrt{\frac{1}{25}\left(\frac{1}{2}\right)}$
$=\pm\frac{1}{5}\left(\frac{\sqrt{2}}{2}\right)$
$=\pm\frac{\sqrt{2}}{10}$
But which sign do I choose?
 A: @N F Taussig has set me straight on this...
$\arcsin$ takes values in $[-\frac{\pi}2,\frac{\pi}2]$, so, in fact, $-\frac{\pi}2\lt \beta\lt0$ (since $\sin{\beta}\lt0$)
So we conclude that $$-\frac{\pi}4\lt\frac{\beta}2\lt0$$ and $\sin{\frac{\beta}2}\lt0$.
A: Let $\sin^{-1}\left(-\frac{7}{25}\right)=t$
As for real $x,-\dfrac\pi2\le\sin^{-1}x\le\dfrac\pi2,-\dfrac\pi2\le t<0$ as $t<0$
$\implies\sin\left(\dfrac t2\right)<0$
and $\cos t=+\sqrt{1-\sin^2t}=?$
$\sin\left(\dfrac t2\right)=-\sqrt{\dfrac{1-\cos t}2}=?$
A: The other answers are fine, but I thought it might help to have an image of what's going on here.  In order to understand this properly, you need to have (at least) a notion of sine and cosine as functions involving the unit circle.

The original angle $\beta$ is determined by when the line $y = -\frac{7}{25}$ intersects the unit circle in the right half-plane ($x \geq 0$, or equivalently $-\frac\pi2 \leq \beta \leq \frac\pi2$).  This is the dashed line and angle in blue.  Half that is the angle in orange, and the desired sine is the dotted line in orange.
In short, since $\beta$ is negative, so must $\beta/2$ be.
