How do you simplify this equation with Sin and Cos

Question

I was solving a Physics problem and I saw someone explained the way to solve it like this:

Eqn.(1)=Eqn.(2)

$$ \frac{2}{6.8cos θ} =\frac{6.8sin θ}{4.9}$$

$$2sin θ cos θ=\frac{4(4.9)}{6.8^2}$$

$$sin2 θ=4(4.9)/(6.8)^2=0.4239$$

$$2θ= 25.81 deg. (or) 180-25.81=154.19 deg$$

$$θ = 12.54 deg. (or) 77.46 deg$$

Reference https://www.physicsforums.com/threads/projectile-motion-angles-question.409546/

So I don't get how he goes from: $$ \frac{2}{6.8cos θ} =\frac{6.8sin θ}{4.9}$$ to $$2sin θ cos θ=\frac{4(4.9)}{6.8^2}$$

How does he get 2sin θ cos θ? If I would have to solve for θ I would cross multiply the fractions, but that doesn't give me what they explain. Does anyone know what I am doing wrong?

Answer 1

$$\frac{2}{6.8\cos(\theta)} =\frac{6.8\sin(\theta)}{4.9}\Longleftrightarrow2\times4.9=6.8\times\cos(\theta)\times6.8\times\sin(\theta)\Longleftrightarrow$$ $$2\times4.9=6.8^2\times\cos(\theta)\times\sin(\theta)\Longleftrightarrow\frac{2\times4.9}{6.8^2}=\cos(\theta)\times\sin(\theta)\Longleftrightarrow$$ $$\frac{2\times4.9}{6.8^2}=\frac{\sin(2\theta)}{2}\Longleftrightarrow\sin(2\theta)=2\times\frac{2\times4.9}{6.8^2}=\frac{2^2\times4.9}{6.8^2}$$

Now, we get two solutions (in radians):

1. $$2\theta=\pi-\arcsin\left(\frac{2^2\times4.9}{6.8^2}\right)+2\pi\text{k}_1\Longleftrightarrow\theta=\frac{1}{2}\left(\pi-\arcsin\left(\frac{2^2\times4.9}{6.8^2}\right)+2\pi\text{k}_1\right)$$
2. $$2\theta=\arcsin\left(\frac{2^2\times4.9}{6.8^2}\right)+2\pi\text{k}_2\Longleftrightarrow\theta=\frac{1}{2}\times\arcsin\left(\frac{2^2\times4.9}{6.8^2}\right)+\pi\text{k}_2$$

Where $\text{k}_1\space\wedge\space\text{k}_2\in\mathbb{Z}$

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Using that:

$$\cos(\theta)\sin(\theta)=\frac{\sin(2\theta)}{2}$$