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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.

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    $\begingroup$ Looks good. You just have to justify why you chose the positive sign for the $\cos \theta$. $\endgroup$
    – Azlif
    Sep 11, 2021 at 1:14

2 Answers 2

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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.

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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.

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