First, use the identity `sin(a + b) = sin(a) cos(b) + cos(a) sin(b),` which is true for any complex `a, b.` We obtain

`sin(pi/2 + i ln(2)) = sin(pi/2) cos(i ln(2)) + cos(pi/2) sin(i ln(2)).`

It is equal to `cos(i ln(2))` because `cos(pi/2) = 0` and `sin(pi/2) = 1.`

Now recall that `cos(z) = (e^(iz) + e^(-iz))/2,` therefore `cos(i x) = (e^(-x) + e^x)/2` and finally

`cos(i ln(2)) = (e^(-ln(2)) + e^ln(2))/2= (1/(e^(ln(2))) + e^ln(2))/2 = (1/2 + 2)/2 = 5/4.`

This is the answer. `e^ln(2) = 2` by the definition of natural logarithm.

We must use the following relationship for a complex number `z` .

`sin(z)=(e^(iz)-e^(iz))/(2i)`

`sin(pi/2+i ln(2))=sin(z)=(e^(i(pi/2+i ln(2)))-e^(i(pi/2+i ln(2))))/(2i)`

`=(e^(i(pi/2))*e^(-ln(2))-e^(-i(pi/2))*e^(ln(2)))/(2i)`

`=(i*1/2-(-i)*2)/(2i)`

`sin(z)=(e^(iz)-e^(iz))/(2i)=5/4`

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