Evaluate $\sin\left(-\frac{\pi}{6} + \frac{1}{2}\arccos\left(\frac{1}{3}\right)\right)$. My task is to evaluate $$\sin\left(-\frac{\pi}{6} + \frac{1}{2}\arccos\left(\frac{1}{3}\right)\right).$$
I think I've gotten most of the way there but I keep running into trouble... any suggestions?
 A: let $$\cos^{-1} \left(\frac13 \right) = t.$$  then we know that $$0 < t < \pi/2, \cos t = 1/3, 
\cos (t/2) = \sqrt{(1+\cos t)/2 } = \sqrt{2/3}, \sin(t/2)=\sqrt{1/3}$$ now we can evaluate $$\sin(-\pi/6 + t/2) = \sin(t/2)\cos(\pi/6) - \cos(t/2)\sin(\pi/6) = \sqrt{1/3}\sqrt{3/4}-\sqrt{2/3}\sqrt{1/4}=\frac12 - \frac1{\sqrt6}  $$
A: A hint:
$$\cos\bigl(\arccos{1\over3}\bigr)={1\over3},\qquad \sin\bigl(\arccos{1\over3}\bigr)={2\over3}\sqrt{2}\ .$$
A: Let $\dfrac12\arccos\dfrac13=y\iff\arccos\dfrac13=2y\implies0<2y<\dfrac\pi2$ and $\cos2y=\dfrac13$
and $0<y<\dfrac\pi4\implies\sin y,\cos y>0$
Use $\cos2y=2\cos^2y-1=1-2\sin^2y$ to find $\sin y,\cos y$
Finally, $\sin\left(-\dfrac\pi6+y\right)=\sin\left(y-\dfrac\pi6\right)=\sin y\cos\dfrac\pi6-\cos y\sin\dfrac\pi6$
A: We very well know $$\cos\theta=2\cos^2\frac{\theta}{2}-1$$ $$\implies \cos\frac{\theta}{2}=\pm \sqrt{\frac{1+\cos\theta}{2}}$$ If $\theta\leq\pi$ $$\implies \cos\frac{\theta}{2}=\sqrt{\frac{1+\cos\theta}{2}} $$
Now, let $\cos^{-1}\frac{1}{3}=\theta $ $ (\theta<\pi)$ or $\cos\theta=\frac{1}{3}$ then we have $$\implies \cos\frac{\theta}{2}=\sqrt{\frac{1+\frac{1}{3}}{2}}=\sqrt{\frac{2}{3}}$$ $$\frac{\theta}{2}=\cos^{-1}\left(\sqrt{\frac{2}{3}}\right)\implies \frac{1}{2}\cos^{-1}\left(\frac{1}{3}\right)=\cos^{-1}\left(\sqrt{\frac{2}{3}}\right)$$ Hence, we have $$\sin\left(-\frac{\pi}{6}+\frac{1}{2}\cos^{-1}\left(\frac{1}{3}\right) \right)=\sin\left(-\frac{\pi}{6}+\cos^{-1}\left(\sqrt{\frac{2}{3}}\right)\right)$$ $$=\sin\left(-\frac{\pi}{6}\right)cos\left(\cos^{-1}\left(\sqrt{\frac{2}{3}}\right)\right)+\cos\left(-\frac{\pi}{6}\right)\sin\left(\cos^{-1}\left(\sqrt{\frac{2}{3}}\right)\right)$$ $$=\left(-\frac{1}{2}\right)\left(\sqrt{\frac{2}{3}}\right)+\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{\sqrt{3}}\right)$$ $$=\frac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}$$
