<s>[INST] How to solve $\sin(2\theta)$ questions Given that: $\sin\theta=\displaystyle{}\frac{12}{13}$ and $0<\theta<\displaystyle{}\frac{\pi}{2}$ the value of $\sin(2\theta)$ is:
I figured out a way to solve it, though I'm not sure if it is the best solution.
Here we will combine two different trigonemtric identities. First:
$\begin{align}
\sin(2\theta) & = 2\sin\theta\cos\theta \\
& = 2\cdot\frac{12}{13}\cdot\cos\theta \\
& = \frac{24}{13}\cdot\cos\theta
\end{align}$
Also:
$\begin{align}
1 & = \cos^2\theta+\sin^2\theta \\
1 & = \cos^2\theta+\Bigg(\frac{12}{13}\Bigg)^2 \\
1 & = \cos^2\theta+\frac{144}{169} \\
1-\frac{144}{169} & = \cos^2\theta \\
\frac{25}{169} & = \cos^2\theta \\
\sqrt\frac{25}{169} & = \sqrt{\cos^2\theta} \\
\frac{5}{13} & = \cos\theta \\
\end{align}$
Then we insert this into the previous equation:
$\begin{align}
\sin(2\theta) & = \frac{24}{13}\cdot\cos\theta \\
& = \frac{24}{13}\cdot\frac{5}{13} \\
& = \frac{120}{169}
\end{align}$
And I believe this is the correct answer. I'm just not sure if this was a super round about way of solving it or if there is something better.
 [/INST] The line
$$\cos^2\theta = \frac{25}{169}$$
simplifies to
$$|\cos\theta| = \frac{5}{13}$$
since $\sqrt{x^2} = |x|$.  Since $\cos\theta > 0$ if $0 < \theta < \dfrac{\pi}{2}$, $|\cos\theta| = \cos\theta$ in this interval, which allows you to conclude that
$$\cos\theta = \frac{5}{13}$$
The rest of your work is correct.
</s>