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The value of $\sin \left[2 \cos ^{-1} \frac{\sqrt{5}}{3}\right]$ is
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Verified Answer
The correct answer is:
$\frac{4 \sqrt{5}}{9}$
We have,
$$
\begin{aligned}
\sin \left(2 \cos ^{-1} \frac{\sqrt{5}}{3}\right) &=\sin \left\{\cos ^{-1}\left(2 \times\left(\frac{\sqrt{5}}{3}\right)^{2}-1\right)\right\} \\
&=\sin \left\{\cos ^{-1}\left(\frac{10}{9}-1\right)\right\} \\
&=\sin \left(\cos ^{-1} \frac{1}{9}\right)
\end{aligned}
$$
$$
\begin{aligned}
&=\sin \left\{\sin ^{-1} \sqrt{1-\left(\frac{1}{9}\right)^{2}}\right\} \\
&=\sin \left\{\sin ^{-1}\left(\frac{4 \sqrt{5}}{9}\right)\right\}=\frac{4 \sqrt{5}}{9}
\end{aligned}
$$
$$
\begin{aligned}
\sin \left(2 \cos ^{-1} \frac{\sqrt{5}}{3}\right) &=\sin \left\{\cos ^{-1}\left(2 \times\left(\frac{\sqrt{5}}{3}\right)^{2}-1\right)\right\} \\
&=\sin \left\{\cos ^{-1}\left(\frac{10}{9}-1\right)\right\} \\
&=\sin \left(\cos ^{-1} \frac{1}{9}\right)
\end{aligned}
$$
$$
\begin{aligned}
&=\sin \left\{\sin ^{-1} \sqrt{1-\left(\frac{1}{9}\right)^{2}}\right\} \\
&=\sin \left\{\sin ^{-1}\left(\frac{4 \sqrt{5}}{9}\right)\right\}=\frac{4 \sqrt{5}}{9}
\end{aligned}
$$
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