I use a bike to commute, so I spend a lot of time thinking about how to get the most bang out of my momentum.
Aside from the extra distance traveled in a wide turn, does making a sharp turn save you any energy? My guess is no, because these things tend to even out, but it definitely feels like I'm going much faster. I've even considered that maybe taking a turn sharp is worse, because the extra pressure will cause more friction in the bearings and the tires.
Either way it's more fun, though that probably doesn't get a term in the equations.
Answer
So this will take a really simplistic look at it, ignoring things like flexibility in the tires/wheels/bike and assuming that you don't go too fast to slide out.
The work done to turn is the force to turn times the distance of the turn. The force is $mv^2/R$ where $m$ is the mass of the system, $v$ is the speed of the turn, and $R$ is the radius of the turn. So the force gets higher as either the speed increases or the turn gets smaller, as expected. The distance of the turn is the arc-length of a circle (assumption of course) of the radius, which is $d = \theta R$.
This means the work done is
$$W = \frac{mv^2}{R} \theta R$$
Or, in other words, if you aren't going to slide out and you can generate the forces you need to make the turn, it doesn't actually matter what radius you choose, the work done is the same either way.
Now this is where the assumptions become obviously violated. We can't turn instantly ($R = 0$) because there isn't enough friction to keep us upright. And the amount of friction increases as the turn gets sharper. The turning friction dominates the friction along the arc-length (unless your tires are slipping riding in a straight line), so all of this comes together to imply that you save energy by taking a bigger turn faster than a narrower turn slower.
But you obviously want to take the tightest turn you can at the fastest speed you can without crashing to maximize everything.
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