Alright, so here we go. The first thing you need are the orbital elements of the planets, particularly earth and Venus. We shall simplify things down to "back of the envelope" calculations which will get us within a fraction of a degree of the right answer. We won't concern ourselves with all the planetary perturbations calculations, it would take me all night to type all that out, instead we'll focus on the mean orbital elements of the planets freed of perturbations for the epoch 1990.0. Here is a scan of a page containing those elements from the book "Practical Astronomy with Your Calculator" by Peter Duffett-Smith.
[link to dropcanvas.com]
The rest is just math. First we need to find the number of days since the epoch of those elements. In fact we can find the apparent position of Venus for a given day and time by converting to fractions of a day since epoch 1990.0. Let's use last night for this demonstration, say at 23:00 UT, which was just after sunset for my location. So October 22, 2013 23:00 UT. Let's convert this date and time to the Julian day number.
y = year
m = month
d = day (fraction of a day)
So here we have
y = 2013
m = 10
d = 22.95833
The procedure to convert this to Julian day number format is as follows:
If m = 1 or 2, subtract 1 from y and add 12 to m, otherwise y' = y and m' = m
If the date is > or = 1582 October 15 calculate:
A = the integer part of y'/100
B = 2-A+integer part of A/4
otherwise B = 0
if y' is negative calculate C = integer of ((365.25 x y')-.75)
otherwise C = integer part of (365.25 x y')
D = integer of (30.6001 x (m' + 1))
JD = B + C + D + d + 1720994.5
In this case JD = 2456588.45833333
We also want the JD for 1990.0, which equals 2447891.5. Therefore, the days since epoch 1990.0 which we will call D, equals 8696.9583333335.
D = 8696.9583333335
We will first calculate the position of earth at this timepoint before we do the calculations for Venus.
Mean anomaly = 360/365.242191 * D/Tp + E - W
where
D = days since epoch 1990.0
Tp = Period
E = Longitude at epoch (1990.0)
W = Longitude of perihelion
Refer to the above scanned page for all of these values for both Earth and Venus. In this case that equates to 8568.4260132099 degrees. Now obviously that's greater than 360, so you need to reduce it down to a value within 360 degrees using the following formula:
d = starting value (8568.4260132099 in this case)
(d/360 - integer of (d/360))*360 = d'
if d' < 0, then 360 + ((d/360 - integer of (d/360))*360) = d'
d' is the new value. This formula will be reused later on, I will simply refer to it as the "d' routine."
Mean anomaly M for earth (which I will call Me) then = 288.4260132098 degrees
Me = 288.4260132098
Next is the true anomaly, v. True anomaly is the angle between the real planet's position and its perihelion position. Normally this calls for an iterative process to solve Kepler's equation, but we will simplify things here by approximating the value with the equation of the center.
v = M + 360/pi * e * sin(M)
where
M = mean anomaly
e = eccentricity
We know the mean anomaly for earth above and eccentricity for earth = 0.016713. Therefore, earth's true anomaly (which I will call ve) equals 286.6090301972 degrees.
ve = 286.6090301972
Now we need to calculate heliocentric longitude, l. For earth I will call it Le.
l = (360/365.242191 * D/Tp) + 360/pi * e * sin(360/365.242191 * D/Tp + E - W)
where
D = days since 1990.0
Tp = Period
e = eccentricity
E = longitude at epoch 1990.0
W = longitude of the perihelion
Le = 8669.3774431972
Use the d' routine to get Le = 29.3774431972 degrees
Now we need the radius vector, that is to say, distance from the sun, r. For earth I will call this re.
r = (a*(1-e^2))/(1+e*cos(v))
where
a = semi-major axis of the orbit
e = eccentricity
v = true anomaly
re = 0.9949674832 AU
Now we repeat these calculations for Venus given the values from the table for Venus.
Mean anomaly for Venus = Mv = 210.6743831713 degrees
True anomaly for Venus = vv = 210.2781423101 degrees
heliocentric longitude for Venus = Lv = 341.7083783101 degrees
radius vector for Venus = rv = 0.7275574534 AU
Now we need the heliocentric latitude for Venus. This is given by
U = Arcsin(sin(Lv - Omega)*sin(i))
where
Lv = heliocentric longitude of Venus
omega = longitude of the ascending node of Venus
i = inclination of Venus
U = -3.3822083713 degrees
Now we need l' and r' which are the heliocentric longitude and radius vector of Venus respectively projected onto the plane of the ecliptic.
l' = arctan((sin(Lv-Omega)*cos(i))/cos(Lv-omega))+omega
Now you need to evaluate the numerator and denominator within the arctan function separately in order to do a quadrant disambiguation routine. If the numerator is positive and the denominator negative, add 180 to the result of the arctan function. If the numerator is negative and the denominator positive, then add 360 to the arctan result unless that makes it greater than 360 degrees, in which case add 180 degrees. Finally, if both the numerator and denominator are negative, then add 180 degrees to the result of the arctan function. Otherwise if you have made no other additions and the result is negative, add 180 degrees, or else keep the result of the arctan function without adding or subtracting anything.
Following this routine, the result you should get for l' is 341.7048475079 degrees.
l' = 341.7048475079 degrees
r' is a simple one, it is simply rv * cos(U)
where
rv = radius vector of Venus
U = heliocentric latitude of Venus
r' = 0.7262901907 AU
Now we calculate the geocentric ecliptic longitude (lambda) and latitude (beta) of Venus. Since Venus is an inner planet the formula is this:
lambda = 180 + Le + arctan ((re*sin(l' - Le))/(re*sin(l'-Le)))
where
Le = heliocentric longitude of earth
re = radius vector of earth
l' = heliocentric longitude of venus projected onto ecliptic
Note that you need to follow the same quadrant disambiguation routine as listed above for the arctan function in this formula. The result in this case should yield
lambda = 256.0825668728 degrees
beta is calculated with the following formula:
beta = arctan((r'*tan(U)*sin(lambda-l'))/(re*sin(l'-Le))
where
Le = heliocentric longitude of earth
re = radius vector of earth
l' = heliocentric longitude of venus projected onto ecliptic
U = heliocentric latitude for Venus
r'= radius vector of Venus projected onto the plane of the ecliptic
Note that you need to follow the same quadrant disambiguation routine as listed above for the arctan function in this formula. The result in this case should yield
beta = -3.3298312039 degrees
For the final step we will need the obliquity of the ecliptic. This can be calculated with the following formula, which will need time "t" centuries from 2000.0. We can calculate t thusly:
t = ((JD-2451545)/36525)
where JD = julian day number
t = 0.1380823637
Obliquity of the ecliptic (Obl) can then be calculated with this formula:
Obl = 23.43928-0.013*t+0.555*(10^-6)*(t^3)-0.0141*(10^-8)*(t^4)
Obl = 23.43748
Now we just have to convert these coordinates from ecliptic to equatorial and we will have the geocentric equatorial coordinates of Venus at this timepoint.
Right ascension = arctan((sin(lambda)*cos(Obl)-tan(beta)*sin(Obl))/cos(lambda))
where
lambda = geocentric ecliptic longitude
beta = geocentric ecliptic latitude
obl = obliquity of the ecliptic
Note that you need to follow the same quadrant disambiguation routine as listed above for the arctan function in this formula. The result in this case should yield
right ascension = 253.8713410337 degrees. If you convert this to hours, minutes, seconds you get 16hr 55m 29.12s
declination = arcsin(sin(beta)*cos(obl)+cos(beta)*sin(obl)*sin(lambda))
where
lambda = geocentric ecliptic longitude
beta = geocentric ecliptic latitude
obl = obliquity of the ecliptic
declination = -26.0216957613 degrees. If you convert this to degrees, minutes, seconds you get -26d 1' 18.1"

So after all that, here are the calculated coordinates of Venus, which is accurate to within a few arcminutes in right ascension and about a quarter of a degree in declination (primarily because we didn't calculate for perturbations from the planets).
RA: 16hr 55m 29.12s
Dec: -26d 1' 18.1"
These are geocentric coordinates, but topocentric parallax will be insignificant at the distance of Venus, at least for naked eye observations like this (it only amounts to a dozen arcseconds or so, which you can't see by eye) so this is good enough as a general case for all observers. Now we just need to figure out which constellation contains these coordinates. I'm going to go to the university library and track down a star atlas. I'll use a 1991 copy of the Cambridge star atlas 2000.0 if it's not checked out (2000.0 is the epoch the atlas is in, not the date of publication). I'll shoot a video of it and upload it later tonight, but for now here are the calculations. Enjoy! :)