# $\sin(2\arcsin(\frac{1}{3}))=?$ 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?

• Why do you think this is not the correct answer? – kingW3 Jun 23 '17 at 21:20
• – jim Jun 23 '17 at 21:20
• @kingW3 The person found $\cos \theta = \sqrt{10}/3 > 1$ – jim Jun 23 '17 at 21:22
• Because there wasn't such ōne. – user36339 Jun 23 '17 at 21:22
• @jim Yeah I agree on that one but,he said bottom that $\frac{2\sqrt{8}}9$ is not an answer. – kingW3 Jun 23 '17 at 21:22

## 2 Answers

Your mistake is that $\cos\theta=\dfrac{2\sqrt{2}}{3}$, not $\dfrac{\sqrt{10}}{3}$.

\begin{eqnarray} \cos\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)\\ &=&2\left(\frac{1}{3}\right)\cdot\frac{2\sqrt{2}}{3}\\ &=&\frac{4\sqrt{2}}{9} \end{eqnarray}

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