Say I have an two parametric equations: $$x(t) = 3 - 2\sin(t) + 3 \cos(t); y(t) = 4 - 3\sin(t) + 2\cos(t)$$
I tried to combine the sin and cos using $$a \sin{(x)} + b\cos{(x)} = \sqrt{a^2+b^2} * \cos{(x + \arctan{(\frac{-b}{a})})}$$
In an attempt to use $\sin(t)^2 + \cos(t)^2 = 1$
But I got to a point where my sin and cos couldn't be combined since their parameters were different. How should I approach this?
I think it makes more sense to solve for the system in terms of $\sin t$ and $\cos t$; then use the circular identity. In particular,
$$\begin{align} x &= 3 - 2 \sin t + 3 \cos t \\ y &= 4 - 3 \sin t + 2 \cos t \\ \end{align}$$
implies
$$\begin{align} 3(-2 \sin t + 3 \cos t &= x-3) \\ 2(-3 \sin t + 2 \cos t &= y-4) \\ \end{align}$$
and their difference eliminates $\sin t$:
$$(9 - 4) \cos t = 3(x-3) - 2(y-4),$$
or $$\cos t = \frac{3x - 2y - 1}{5}. \tag{1}$$
Similarly, $$\begin{align} 2(-2 \sin t + 3 \cos t &= x-3) \\ 3(-3 \sin t + 2 \cos t &= y-4) \\ \end{align}$$
and this time, their difference eliminates $\cos t$:
$$(-4 + 9) \sin t = 2(x-3) - 3(y-4),$$
or $$\sin t = \frac{2x-3y+6}{5}. \tag{2}$$
Therefore, $$\cos^2 t + \sin^2 t = 1$$ gives us
$$(3x - 2y - 1)^2 + (2x - 3y + 6)^2 = 5^2,$$
which is a conic section; expressed in standard form, it is
$$13x^2 - 24xy + 13y^2 + 18x - 32y + 12 = 0. \tag{3}$$