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`
See eNotes Ad-Free
Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.
Already a member? Log in here.
Further Reading