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

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Borys Shumyatskiy eNotes educator| Certified Educator

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.

Charlie Hooper eNotes educator| Certified Educator

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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