How do they derive these formulas? I've been doing some practice problems and they use formulas such as
$$x=\frac{v_0^2\sin(2\theta)}{g}$$
$$\Delta y=\frac{v^2\sin^2(\theta)}{2g}$$
Can somebody please inform me on the ways in which they derive these formulas?
 A: The first equation is for the projectile range, $R$,
$$R=\frac{v_0^2\sin(2\theta)}{g}$$
This assumes that the projectile starts and ends at the same height.
We know that $\Delta y = -\dfrac12 gt^2+v_0 \sin(\theta) t$ where $v_{0,\; y}=v_0\sin\theta$. Since the projectile lands at the same height from which it was launched, we have $\Delta y = 0$, which means: $$0=-\dfrac12 gt^2 + v_0\sin(\theta) t\quad \implies 0=t\left(-\dfrac12gt+v_0\sin\theta\right).$$
The two times $t$ for which $\Delta y=0$ are $t_1=0$ and $t_2=\dfrac{2v_0 \sin\theta}g$. The trivial solution is $t=0$, we only care about the landing time, $t_2=\dfrac{2v_0 \sin\theta}g$.
Recall that the x-displacement is $\Delta x = v_0\cos\theta \cdot t$. We're after the range, so we get $$R=v_0\cos\theta \cdot t_2\implies R=v_0 \cos\theta \dfrac{2v_0 \sin\theta}g.$$
Recall the trig identity $\sin2\theta \equiv 2\sin\theta\cos\theta$. Applying it, we get,
$$R=\dfrac{v_0^2 \sin2\theta}{g}.$$

The second equation is for the max displacement of your projectile, $$\Delta y_\mathrm{max}=\frac{v^2\sin^2(\theta)}{2g}.$$
Easiest way to get it is to use the $v_f^2 = v_0^2 +2a\Delta d$ kinematic equation. Note that we are working in the y-direction, so this equation becomes $$v_{f, \ y}^2 = v_{0, \ y}^2 -2g\Delta y.$$
Note that the y-velocity at the max height is zero; that's what makes the height a max. We then have, $$2g\Delta y = v_{0, \ y}^2.$$
We know that $v_{0, \ y} = v_0 \sin\theta$, so we have
$$2g\Delta y = \left(v_0 \sin\theta \right)^2$$
which gives the max displacement and simplifies to $$\Delta y_\mathrm{max}=\frac{v^2\sin^2(\theta)}{2g}.$$
A: Thank you to everybody who commented and tried helping, I found out the answer. For the $\Delta x$ formula you use $v_{fy} = v_{0y} + ay \cdot t$ and then you isolate $t$ and plug it into $s_x = v_x \cdot t$ then use pythagorean identities and you have the first formula!
For the next one use $v_f^2 = v_0^2 + 2ad$ then isolate $y$ granted that $v_{fy} = 0$. Then you have your second formula!
A: Nothing wrong with those two formulas.
The distance traveled by any particle along a direction can be written using the following two formulas. This you can find in any basic physics book.
\begin{align}
S_{x} &=u_x*T ~~~~\text{when moving at constant velocity} \label{eq:01} \\
S_y &=u_y*t+\frac{1}{2}f*T^2 ~~~~\text{when moving with acceleration}  \label{eq:02} \\
\end{align}
In your problem, for the horizontal direction moion,
\begin{align}
S_{x} &=u_x*t  \label{eq:03} \\
S_{x} &=v_0*cos(\theta)*T  \\
T &= \frac{S_{x}}{v_0cos(\theta)} \\
\end{align}
For the verticle motion, $S_y$ is zero as the particle starts from the ground then goes up certain distance and then comes back to ground, net verticle distance is zero.
\begin{align}
S_y &=u_y*T+\frac{1}{2}f*T^2 \label{eq:04} \\
0 &= v_0sin(\theta)*T + \frac{1}{2}*(-g)*T^2  \\
~~~~ & \text{Note: g is acting in -ve y direction, that's why -ve sign}\\
v_0sin(\theta)*T &= \frac{1}{2}*g*T^2 \\
T &= \frac{2v_0sin(\theta)}{g} \\
\end{align}
Equate those two T's you will get your first equation.
\begin{align}
\frac{S_{x}}{v_0cos(\theta)} &= \frac{2v_0sin(\theta)}{g} \\
S_x &= \frac{v_0^2sin(2\theta)}{g}\\
\end{align}
\begin{equation}
\boxed{x = \frac{v_0^2sin(2\theta)}{g}}
\end{equation}
$\Delta{y}$ is the maximum height. At max verticle height $v_y=0$. Use this condition to get the time it takes to reach maximum height.
\begin{align}
v_y &= v_0sin(\theta) -g*T' ~~~\text{basic formula: v=u+ft}\\
0 &= v_0sin(\theta) -g*T' \\
T' &= \frac{v_0sin(\theta)}{g} \\
\end{align}
Then maximum distance traveled in the verticle direction becomes,
\begin{align}
\Delta{y} &= v_0sin(\theta)*T' -\frac{1}{2}g*{T'}^2 \\
          &= v_0sin(\theta)*\left[\frac{v_0sin(\theta)}{g} \right] -\frac{1}{2}g*{\left[ \frac{v_0sin(\theta)}{g} \right] }^2 \\
          &= \frac{v_0^2sin^2(\theta)}{2g} \\
\end{align}
\begin{equation}
\boxed{\Delta{y} = \frac{v_0^2sin^2(\theta)}{2g}}
\end{equation}
