Euler's Identity states that e^(ix)=cosx+isinxeix=cosx+isinx, and from that e^((23pi)/12i)=cos((23pi)/12)+isin((23pi)/12)e23π12i=cos(23π12)+isin(23π12) and e^((13pi)/2i)=cos((13pi)/2)+isin((13pi)/2)e13π2i=cos(13π2)+isin(13π2).
Substituting these into your problem, we have cos((23pi)/12)+isin((23pi)/12)-cos((13pi)/2)-isin((13pi)/2)cos(23π12)+isin(23π12)−cos(13π2)−isin(13π2). If you can use a calculator then you can stop here to find the values of these functions.
If you need an exact answer there's a bit farther to go in order to put these functions in terms of special angles.
Since sin(theta)=sin(theta+2npi)sin(θ)=sin(θ+2nπ) and cos(theta)=cos(theta+2npi)cos(θ)=cos(θ+2nπ) where ninZZ we can say n="-"6 and change our expression to cos((23pi)/12)+isin((23pi)/12)-cos((pi)/2)-isin((pi)/2)
Since sin(2npi-theta)=sin("-"theta)="-"sin(theta) and cos(2npi-theta)=cos("-"theta)=cos(theta) where ninZZ we can say n=1 and change our expression to cos((pi)/12)-isin((pi)/12)-cos((pi)/2)-isin((pi)/2).
Finally, since pi/12=(pi/6)/2, our expression is cos((pi/6)/2)-isin((pi/6)/2)-cos((pi)/2)-isin((pi)/2) and we can use the half-angle formulas sin(theta/2)=+-sqrt((1-cos(theta))/2) and cos(theta/2)=+-sqrt((1+cos(theta))/2) leaving us with +-sqrt((1+cos(pi/6))/2)-i(+-sqrt((1-cos(pi/6))/2))-cos((pi)/2)-isin((pi)/2). Ugly, I know, but easy since cos(pi/6)=sqrt3/2, cos(pi/2)=0, and sin(pi/2)=1.
"+"sqrt((1+sqrt3/2)/2)-i("+"sqrt((1-sqrt3/2)/2))-0-i so the answer is sqrt((1+sqrt3/2)/2)-isqrt((1-sqrt3/2)/2)-i
