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

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Tom Share
Feb 21, 2017

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)#

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