Find the value of sin(pi/2+i ln(2)) .

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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