If you’re standing at a latitude of $theta^{,circ}$, then you’re rotating on a circle of radius $Rcos(theta)$, where $R = 3959,mathrm{mi}$ is the radius of the earth. This circle still does one revolution in (about!) $24,mathrm{hrs}$, so we can calculate the speed from there.

$$

left(frac{1}{24 ,mathrm{hours}}right) times 2pi Rcostheta ,mathrm{miles}

approx 1036costheta ,mathrm{mph}

$$

So you’ve just gotta scale the speed at the equator by $costheta$.