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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

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

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

Wolfram Alpha is a good tool for checking such things.

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  • $\begingroup$ Thanks, Khallil, I'll check that tool out! $\endgroup$ – CopperKettle Aug 7 '15 at 11:51
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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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  • $\begingroup$ 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.. $\endgroup$ – CopperKettle Aug 7 '15 at 14:21

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