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# prove the following trigonometric equation?

Asked by Jagmeet Singh 13th October 2013, 12:31 PM

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

sin(2?)+sin(2?)+sin(2?)=0
Answered by Expert 13th October 2013, 4:11 PM
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