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f[t_] = -4 + Sqrt[(-Sqrt[3] + 2*Cos[t])^2 + Sin[t]^2] + Sqrt[(Sqrt[3] 
+ 2*Cos[t])^2 + Sin[t]^2] 

The function above is identically 0 for all real values of t. However:

FullSimplify[f[t],Reals] 

just spits back the original function.

I also tried TrigReduce, TrigExpand, etc, to no avail.

How do I make Mathematica simplify things like this?

For those interested: I'm playing w/ ellipses, and f[t] represents the sum of the distance from the focii minus the actual value of the sum of the distance from the focii. The more general case:

x[t_] = a*Cos[t]  
y[t_] = b*Sin[t]  
focus[a_,b_] = Sqrt[a^2-b^2]  
f[t_]=Sqrt[(x[t]-focus[a,b])^2 + y[t]^2] + Sqrt[(x[t]+focus[a,b])^2 + y[t]^2] 

where f[t] is constant for real t when a >= b (probably when a < b too, but things get a bit complex in that case)

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1 Answer 1

Simplification uses heuristics, so there's no guarantee it'll find the simplest form. As you'll see below, it takes some work to make the simplification possible.

Note: In your example you were using incorrect syntax: the second argument of FullSimplify can't be Reals. It must be a set of assumptions about some variables. Take a look at the examples on the doc page to see how to use assumptions: FullSimplify.

After correcting this mistake, Mathematica still needs some guidance to be able to handle this. Your expression has the sum of two square roots:

expr = -4 + Sqrt[(-Sqrt[3] + 2*Cos[t])^2 + Sin[t]^2] + 
  Sqrt[(Sqrt[3] + 2*Cos[t])^2 + Sin[t]^2]

Let's transform it so it'll only have one square root and see if that helps:

Simplify[Expand[(expr + 4)^2], t \[Element] Reals]

(* ==> 16 *)

Since (expr + 4)^2 is 16 for all real t, expr is either 0 or -8. By substituting in t -> 0 and noting that the function is continuous, we can quickly verify that expr == 0 for all real t.

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