CBSE Class 11-science Answered
Let x= cos(?)+isin(?)
y = cos(?)+isin(?)
z= cos(?)+isin(?)
x+y+z = cos(?)+cos(?)+cos(?)+i(sin(?)+sin(?)+sin(?)
x+y+z = 0 + i 0 =0
(x+y+z)2=0
x2+y2+z2=?2(xy+yz+zx)
x2+y2+z2=?2xyz(1/x+1/y+1/z) -----------(1)
1/x = 1/ cos(?)+isin(?)
1/x=cos(?)?isin(?)
Similarly we get
1/y=cos(?)?isin(?)
1/z=cos(?)?isin(?)
1/x+1/y+1/z=cos(?)+cos(?)+cos(?)?i(sin(?)+sin(?)+sin(?))
=0 - i (0) =0
Therefore from (1), we get
x2+y2+z2=?2xyz(0)
x2+y2+z2=0 ----------(2)
Using De Moivre's Theorem we get
x2=(cos(?)+isin(?))2 = cos(2?)+isin(2?)
Similarly we get
y2=cos(2?)+isin(2?)
z2=cos(2?)+isin(2?)
x2+y2+z2=0
x2+y2+z2=cos(2?)+cos(2?)+cos(2?)+i(sin(2?)+sin(2?)+sin(2?))=0 [From (2)]
Hence,
cos(2?)+cos(2?)+cos(2?)=0