What does #(e^(ix)+e^(ix))/(2i)# equal?

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Answer 1 · 2015-10-25T12:40:15

#sin(x) - i cos(x)#

but I think you meant to ask...

#e^(ix) = cos(x) + i sin(x)#

#cos(-x) = cos(x)#

#sin(-x) = -sin(x)#

So with the question as asked:

#(e^(ix) + e^(ix))/(2i) = e^(ix)/i = (cos(x) + i sin(x))/i = sin(x)-i cos(x)#

I think you may have been wanting one of the following results:

#(e^(ix)+e^(-ix))/2#

#= ((cos(x)+i sin(x)) + (cos(-x)+i sin(-x)))/2#

#= ((cos(x)+i sin(x)) + (cos(x)-i sin(x)))/2#

#= cos(x)#

#color(white)()#

#(e^(ix)-e^(-ix))/(2i)#

#= ((cos(x)+i sin(x)) - (cos(-x)+i sin(-x)))/(2i)#

#= ((cos(x)+i sin(x)) - (cos(x)-i sin(x)))/(2i)#

#= sin(x)#