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)
