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Can I force Mathematica to re-express something like $$\small \left\{\frac{1}{2}i\left(\frac{1}{16} e^{-2 i (\alpha (1)+4 \alpha (2))} \left(-1+e^{8 i \alpha (2)}\right) \text{d$\alpha $}(1) \left(2 \left(2 e^{2 i \alpha (2)}-2 e^{6 i \alpha (2)}+e^{8 i \alpha (2)}+1\right) \cos (\alpha (4))+e^{4 i \alpha (2)} (\cos (2 \alpha (4))+3)\right)\right.\right.$$ as an expression with solely goniometric functions instead of these imaginary exponentials?

1/2 I (1/16 E^(-2 I (α[1] + 4 α[2])) (-1 + E^(
 8 I α[
   2])) (2 (1 + 2 E^(2 I α[2]) - 2 E^(6 I α[2]) + 
    E^(8 I α[2])) Cos[α[4]] + 
 E^(4 I α[2]) (3 + Cos[2 α[4]])) dα[1]
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  • $\begingroup$ Including Mathematica code for this expression may help you get an answer sooner. $\endgroup$ – Mr.Wizard Oct 25 '14 at 11:04
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You have a missing closing bracket, I think.
Adding that, and applying %//ExpToTrig//FullSimplify gives:

$$-\frac{1}{16} e^{-2 i \alpha (1)} \text{d$\alpha $}(1) \sin (4 \alpha (2)) (\cos (2 \alpha (4))+4 \cos (\alpha (4)) (\cos (4 \alpha (2))-2 i \sin (2 \alpha (2)))+3)$$
Remark that the FullSimplify puts a little exponential back in.

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Assuming your expression is:

exp = 1/2 I (1/
     16 E^(-2 I (α[1] + 4 α[2])) (-1 + 
      E^(8 I α[2])) (2 (1 + 2 E^(2 I α[2]) - 
         2 E^(6 I α[2]) + E^(8 I α[2])) Cos[α[
         4]] + E^(4 I α[2]) (3 + Cos[2 α[4]])) )

You can:

cm = ComplexExpand[ExpToTrig[FullSimplify@ExpToTrig[Expand[exp]]]];
re = FullSimplify[
   Refine[Re[
     cm], {α[1] ∈ Reals, α[2] ∈ 
      Reals, α[3] ∈ Reals, α[4] ∈ 
      Reals}]];
im = FullSimplify[
   Refine[Im[
     cm], {α[1] ∈ Reals, α[2] ∈ 
      Reals, α[3] ∈ Reals, α[4] ∈ 
      Reals}]];
re + I im

yields:

1/16 I ((3 + 4 Cos[4 α[2]] Cos[α[4]] + 
       Cos[2 α[4]]) Sin[2 α[1]] + 
    8 Cos[2 α[1]] Cos[α[4]] Sin[2 α[2]]) Sin[
   4 α[2]] - 
 1/16 (Cos[
      2 α[1]] (3 + 4 Cos[4 α[2]] Cos[α[4]] + 
       Cos[2 α[4]]) - 
    8 Cos[α[4]] Sin[2 α[1]] Sin[2 α[2]]) Sin[
   4 α[2]]

but may be a starting point

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