How do you verify the identity #sin(pi/2 + x) = cosx#?

Question

Anna Andrews · Accepted Answer

You must use matrice for the "true" proof, but the following will do: #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 # Thus, we have: #sin(pi/2+x) = cos(x)# Given that the student finds this response to be highly helpful, please see the entire demonstration to #sin(a+b) = sin(a)cos(b)+cos(a)sin(b)# (If math is not your thing, don't read this.) Trigonometric form can be used to express complex numbers. #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# combining the two parts #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 a different complicated 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)# For an imaginary part, the real part of the left must equal the real part of the right. #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)#