High latitude launch sites are better for polar or highly inclined orbits.
Consider a launch from the equator targeting a pure polar orbit (i.e. going over the poles). You not only have to reach orbit, but also cancel out the "equatorial direction" component of your initial velocity to zero, which takes extra delta-v.
Polar orbits are good for a single satellite to see the entire Earth's surface (basically scanning over a different part each orbit), which is often desirable depending on the purpose. For example something that wants to measure the entire atmosphere or photograph the entire surface, etc.
This isn't true. Because it's a vector addition of velocity vectors—not scalars.
The orbit you want to reach is a delta-v of about 9.0 km/s in the polar direction,
↑ 9.0 km/s
The speed of the Earth's rotation, that you would want to cancel, is maximally 0.46 km/s on the equator, in the azimuthal direction,
→ 0.46 km/s
So your delta-v vector is ↑ 9.0 km/s pointing north, plus ← 0.46 km/s pointing west (cancelling the Earth's rotation), for a vector sum of magnitude
sqrt( (9.0 km/s)^2 + (0.46 km/s)^2 )
= 9.01 km/s
Launching into a polar orbit is equally easy, from any latitude!
⁂
The converse isn't true. If you're trying to reach an equatorial orbit, the delta-v into that orbit is co-linear with the earth's rotation,
(→ 9.0 km/s), (→ 0.46 km/s)
That's a scalar sum! An azimuthal launch is 460 m/s cheaper on the equator—you inherit the full speed of the Earth's rotation. Latitude is a significant factor here: it's easier to launch, the closer to you are to the equator.
(More profoundly, the orbit you're trying to reach might not even pass through the point you're trying to launch from. This happens if the target orbit inclination is smaller than your launch site's latitude (in absolute value)).
And they can time it so they get the sun at the same relative location for each photo pass.