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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?

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@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 $\sin{\frac{\beta}2}\lt0$.

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  • $\begingroup$ I read the solution, and it says It's negative, how is this possible? $\endgroup$ – Whoj Jeong Jul 14 '18 at 1:33
  • $\begingroup$ Even if $\cos{\beta}\lt0$, (which it might be, upon closer scrutiny) the $\sin$ comes out positive... so I suspect an error in the book... $\endgroup$ – Chris Custer Jul 14 '18 at 2:14
  • $\begingroup$ @ChrisCuster, The book is right. $\endgroup$ – lab bhattacharjee Jul 14 '18 at 6:31
  • $\begingroup$ Since the range of $f(x) = \arcsin x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, $-\frac{\pi}{2} < \beta = \arcsin\left(-\frac{7}{25}\right) < 0 \implies \cos\beta < 0$ and $-\frac{\pi}{4} < \frac{\beta}{2} < 0 \implies \sin\left(\frac{\beta}{2}\right) < 0$. $\endgroup$ – N. F. Taussig Jul 14 '18 at 8:24
  • $\begingroup$ Ok. My mistake... $\endgroup$ – Chris Custer Jul 14 '18 at 15:38
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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 $\cos t=+\sqrt{1-\sin^2t}=?$

$\sin\left(\dfrac t2\right)=-\sqrt{\dfrac{1-\cos t}2}=?$

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  • $\begingroup$ @WhojJeong, The book is right. $\endgroup$ – lab bhattacharjee Jul 14 '18 at 6:31
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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.

enter image description here

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.

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