# 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:

$$\sin\left(2\arcsin\left(\frac{1}{3}\right)\right)$$

$$\frac{4\sqrt{2}}{9}$$

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

$$2*\frac{1}{3}*\sqrt{1-\frac{1}{9}}=\frac{2\sqrt{8}}{3*3}=\frac{4\sqrt{2}}{9}$$

$$\frac{2\sqrt{2}}{3}$$

• Are you sure that ,the textbook's answer is correct ? – Khosrotash Aug 7 '15 at 11:50
• @Khosrotash - no, that texbook is full of typos, nathless being good in other regards. I'm going to switch to another one. – CopperKettle Aug 7 '15 at 11:50
• I am sure that you do it correct – Khosrotash Aug 7 '15 at 11:52

Notice, $$\sin 2A=2\sin A\cos A$$ & $$|\cos A|=\sqrt{1-\sin^2 A}$$ Now, we have $$\sin\left(2\sin^{-1}\left(\frac{1}{3}\right)\right)$$ $$=\sin\left(\sin^{-1}\left(2\frac{1}{3}\sqrt{1-\frac{1}{3^2}}\right)\right)$$ $$=\sin\left(\sin^{-1}\left(\frac{4\sqrt 2}{9}\right)\right)=\frac{4\sqrt 2}{9}$$
• I'm afraid I don't understand how you arrived to $$\sin\left(\sin^{-1}\left(2\frac{1}{3}\sqrt{1-\frac{1}{3^2}}\right)\right)$$. Did you mean to say that $$2\arcsin(A)=\frac{\arcsin(2A)}{\arccos(A)}?$$ Still that did not result in the same formula in my calculations.. – CopperKettle Aug 7 '15 at 14:21