$\sin(2\arcsin(\frac{1}{3}))=? $ A broad one I have written it as $$\sin(2\arcsin(\frac{1}{3}))=2\sin(\arcsin(\frac{1}{3}))\cos(\arcsin(\frac{1}{3})).$$Then, solved for $$\cos(\arcsin(\frac{1}{3}))=\frac{\sqrt{10}}{3}$$ $$\arcsin(\frac{1}{3})=\theta$$$$\sin\theta=\frac{1}{3}\Rightarrow\cos\theta=\frac{\sqrt{10}}{3}.$$But noticed something weird (1) it didn't satisfy the Pythagorean theorem as $\sqrt{10}\approx3.1622,$ which means that the adjacent side to the angle $\theta$ is greater than the hypotenuse, (which is just 3) why this method (using right triangle to determine $\cos(\theta)$) didn't work here although the answer ($\frac{\sqrt{10}}{3}$) satisfied the range of the cosine function, which is $R(\cos\theta)=\Bbb{R}$? After it (2) used this formula$$\arcsin(x)=\arccos(\sqrt{1-x^2});$$$$\arcsin(\frac{1}{3})=\arccos(\sqrt{1-\frac{1}{9}})=\arccos(\frac{\sqrt{8}}{3});$$then, checked if it satisfies or not $D(\arccos(\theta))=[-1;1],$ satisfied, since $\frac{\sqrt{8}}{3}\approx0.942.$ Then proceeded on my problem $$2\sin(\arcsin(\frac{1}{3}))\cos(\arcsin(\frac{1}{3}))=\frac{2}{3}\cdot\frac{\sqrt{8}}{3}=\frac{2\sqrt{8}}{9}.$$ Anyway, it didn't turn out to be the right answer. Where is my mistake?

$$\sin2\arcsin\frac{1}{3}=2\cdot\frac{1}{3}\cdot\sqrt{1-\frac{1}{9}}=\frac{4}{9}\sqrt2$$
