If sin(x) = -3/4 and x is in the 3rd quadrant, what's sin(x/2)?
I tried subbing in a variable like alpha so
sin(2a)=2cos(a)sin(a)
sin(x)=2cos(x/2)sin(x/2)
Then solving for sin(x/2) but I'm still stuck with the cos part.
You don't HAVE to use the cosine double angle formula, but I do indeed suggest it, as mzhao points out. Otherwise, you're left with working out like this.
\(
\frac{9}{16} = \sin^2(x) = 4\sin^2\left(\frac{x}{2}\right)\cos^2\left(\frac{x}{2}\right)\\
= 4\sin^2\left(\frac{x}{2}\right)\left(1 - \sin^2\left(\frac{x}{2}\right)\right)\\
\sin^4\left(\frac{x}{2}\right) - \sin^2\left(\frac{x}{2}\right) + \frac{9}{64} = 0\\
\sin^2\left(\frac{x}{2}\right) = \frac{1\pm\sqrt{1 - \frac{9}{16}}}{2} = \frac{4\pm\sqrt{7}}{8}
\)
Now, we have a seriously awkward dilemma, and this is why you don't go around squaring things (you introduce extra solutions), as both signs admit valid solutions. As a heuristic, if \(\sin(x) = -0.75\), it is around \(-0.866\), so \(x\approx \frac{4\pi}{3}, \frac{x}{2}\approx \frac{2\pi}{3}\). As such, we expect \(\sin^2\left(\frac{x}{2}\right) \approx \frac{3}{4}\), which warrants taking the positive root as the negative root gives a number closer to 0.5. Indeed, it can be shown that \(\cos^2\left(\frac{x}{2}\right) = \frac{4 - \sqrt{7}}{8}\). Anyway, continuing on and accounting for the quadrant of \(\frac{x}{2}\) gives the same result as mzhao.
Edit: TeX on AN is flat-out not working properly for me for some reason. I had to run the code differently to make it show at all.