for the "true" proof you need to use matrice, but this is acceptable :
#sin(a+b) = sin(a)cos(b)+cos(a)sin(b)#
#sin(pi/2+x) = sin(pi/2)*cos(x)+cos(pi/2)*sin(x)#
#sin(pi/2) = 1#
#cos(pi/2) = 0 #
So we have :
#sin(pi/2+x) = cos(x)#
Since this answer is very usefull for student here the full demonstration to obtain
#sin(a+b) = sin(a)cos(b)+cos(a)sin(b)#
(do not read this if you are not fan of math)
a complex numbers can be written in trigonometrics form
#z = (cos(x) + isin(x))# # -> (1)#
multiplying #z# by #i# you have
#iz = -sin(x) + icos(x)#
because #i^2 = i*i = -1#
just for you to know, multiplying a complex numbers by #i# is the same to do a 90° rotation on the complex plane
another way to do a 90° rotation is to derivate #z#
#z' = -sin(x) + icos(x) #
we have
#z' = iz#
#(z')/z = i#
integrating both part
#ln(z) = ix + C#
#z = e^(ix)e^(C)#
taking #x = 0# and comparing with #(1)# you see that C must be #= 0#
so #z = e^(ix)#
#e^(ix) = cos(x)+isin(x)#
multiplying by another complex number
#e^(ix)e^(ix_0) = (cos(x)+isin(x))(cos(x_0)+isin(x_0))#
#e^(ix)e^(ix_0) = e^(i(x+x_0)#
#e^(i(x+x_0)) = cos(x+x_0)+isin(x+x_0)#
#(cos(x+x_0)+isin(x+x_0) = (cos(x)+isin(x))(cos(x_0)+isin(x_0))#
develop
#(cos(x+x_0)+isin(x+x_0) = cos(x)cos(x_0)+icos(x)sin(x_0) + isin(x)cos(x_0) - sin(x)sin(x_0)#
real part of left must be equal to real part of right idem for imaginary part
#sin(x+x_0) = cos(x)sin(x_0) + sin(x)cos(x_0)#
note :
#sin(x-x_0) = -cos(x)sin(x_0) + sin(x)cos(x_0)#
because #sin(-x)= -sin(x)# and #cos(-x) = cos(x)#
