# 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.

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

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.

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.