# Come si verifica l'identità sin (pi / 2 + x) = cosx ?

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)

moltiplicando z by i hai

iz = -sin(x) + icos(x)

perché 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)

ne ha

z' = iz

(z')/z = i

integrating both part

ln(z) = ix + C

z = e^(ix)e^(C)

presa 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))

sviluppare

(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)

nota:

sin(x-x_0) = -cos(x)sin(x_0) + sin(x)cos(x_0)

perché sin(-x)= -sin(x) e cos(-x) = cos(x)