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.

Another way to do this is to first find $\cos\theta$. This is easier if you recognise small Pythagorean triads. ;)
Let
$y/r=\sin\theta$ and $x/r=\cos\theta$, where $r^2=x^2+y^2$.
We have $\sin\theta=5/13$, so
$$x^2 = 13^2-5^2 = (13+5)(13-5) = 18\cdot8=12^2$$
thus $\cos\theta=12/13$.
Now, we know that $\sin(2\theta)=2\sin\theta\cos\theta$. But that's the middle term of
$$(\sin\theta + \cos\theta)^2 = \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta$$
And of course $\sin^2\theta + \cos^2\theta = 1$
So
$$(\sin\theta + \cos\theta)^2 = 1 + \sin(2\theta)$$
Therefore,
$$\begin{align}\\
\sin(2\theta) & = (5/13 + 12/13)^2 - 1\\
& = (17/13)^2 - 1\\
& = (17^2 - 13^2)/13^2\\
& = 120/169
\end{align}$$

Of course, we need to check the signs of our trig ratios to make sure they're all in the correct quadrant.
