<s>[INST] Am I right in calculating $\sin(2\arcsin(\frac{1}{3}))$ as $\frac{4\sqrt{2}}{9}$? I've been solving a problem in my textbook, and my result is at odds with the textbook's:
$$\sin\left(2\arcsin\left(\frac{1}{3}\right)\right)$$ 
My answer is 
$$\frac{4\sqrt{2}}{9}$$
I've used the double-angle identity for sine. 
$$\sin\left(2\arcsin\left(\frac{1}{3}\right)\right)=2\sin\left(\arcsin\left(\frac{1}{3}\right)\right)\cos\left(\arcsin\left(\frac{1}{3}\right)\right)$$ 
and this identity:
$$\cos(\arcsin(x))=\sqrt{1-x^2},$$
yielding
$$2*\frac{1}{3}*\sqrt{1-\frac{1}{9}}=\frac{2\sqrt{8}}{3*3}=\frac{4\sqrt{2}}{9}$$
But the textbook's answer is 
$$\frac{2\sqrt{2}}{3}$$
 [/INST] Notice, $$\sin 2A=2\sin A\cos A$$ & $$|\cos A|=\sqrt{1-\sin^2 A}$$  Now, we have $$\sin\left(2\sin^{-1}\left(\frac{1}{3}\right)\right)$$ $$=\sin\left(\sin^{-1}\left(2\frac{1}{3}\sqrt{1-\frac{1}{3^2}}\right)\right)$$
$$=\sin\left(\sin^{-1}\left(\frac{4\sqrt 2}{9}\right)\right)=\frac{4\sqrt 2}{9}$$
O.P. is right. There is some probable error in the answer in the textbook.
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