## Trigonometry (11th Edition) Clone

$$\sin x=-\frac{\sqrt{6}}{6}$$
$$\cos2x=\frac{2}{3}\hspace{1.5cm}\pi\lt x\lt\frac{3\pi}{2}\hspace{1.5cm}\sin x=?$$ Though this exercise can certainly be solved using half-angle identities, if you do not want to deal with the square root, another way to do it is through double-angle identities. - Double-angle Identity for cosines which only involves sines: $$\cos2x=1-2\sin^2x$$ $$\sin^2x=\frac{1-\cos2x}{2}$$ $$\sin^2x=\frac{1-\frac{2}{3}}{2}$$ $$\sin^2x=\frac{\frac{1}{3}}{2}$$ $$\sin^2x=\frac{1}{6}$$ Thus, $$\sin x=\pm\frac{\sqrt1}{\sqrt{6}}=\pm\frac{1}{\sqrt6}=\pm\frac{\sqrt6}{6}$$ As $\pi\lt x\lt \frac{3\pi}{2}$, $x$ lies in the area of quadrant III, where sines are negative. So $\sin x\lt0$. Therefore, $$\sin x=-\frac{\sqrt{6}}{6}$$