Finding $\sin 2\alpha$, given $\sin\frac{\alpha}{2}+\cos\frac{\alpha}{2}=-\frac{1}{2}$. (Where's my error?) The problem: > If $$\sin\left(\frac{\alpha}{2}\right)+\cos\left(\frac{\alpha}{2}\right)=-\frac{1}{2},$$ and $\frac{3\pi}{2}\lt \alpha \lt 2\pi,$ find $\sin2\alpha.$ It looks like an ordinary one, so I raised both sides of the first expression to quadrat and had $$\begin{align}\sin^2\left(\frac{\alpha}{2}\right)+\cos^2\left(\frac{\alpha}{2}\right)+2\sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{2}\right)&=\phantom{-}\frac{1}{4}\\\\[6pt] 1+\sin(2\alpha)&=\phantom{-}\frac{1}{4}\\\\[6pt] \sin(2\alpha)&=-\frac{3}{4} \end{align}$$ But there is no such an answer, so any hints concerning this issue will be helpful!

The second last displayed line should read

$$1+\sin \alpha = \frac{1}{4}$$

since you're doubling $\alpha/2$. So $\sin \alpha = -3/4.$ Then you have, since you're in the 4th quadrant and cosine is positive there:

$$\sin 2\alpha = 2\sin \alpha \cos \alpha = 2\frac{-3}{4}\sqrt{1- \left(\frac{-3}{4}\right)^2 }= \frac{-3}{2}\sqrt{\frac{7}{16} }.$$