If cos alpha + cos beta + cos gamma = 0 and alos sin alpha + sin beta + sin gamma= 0, then provethat.(a)cos 2 alpha + cos 2 beta + cos 2gamma = sin 2alpha +sin2beta+sin2gamma=0(b)sin 3 alpha+ sin 3 beta + sin3 gamma = 3 sin (alpha + beta + gamma)(c)cos 3 alpha + cos 3beta + cos 3gamma = 3 cos (alpha + beta + gamma)

If cos alpha + cos beta + cos gamma = 0 and alos sin alpha + sin beta

1.	If cos alpha + cos beta + cos gamma = 0 and alos sin alpha + sin beta + sin gamma= 0, then provethat.(a)cos 2 alpha + cos 2 beta + cos 2gamma = sin 2alpha +sin2beta+sin2gamma=0(b)sin 3 alpha+ sin 3 beta + sin3 gamma = 3 sin (alpha + beta + gamma)(c)cos 3 alpha + cos 3beta + cos 3gamma = 3 cos (alpha + beta + gamma)
Answer» Solution :Let `z_(1) = cos ALPHA + isin alpha,z_(2) = cos beta + isin beta`, `z_(3) = cos gamma +isin gamma` `thereforez_(1) +z_(2)+z_(3) = (cos alpha + cos beta + cos gamma)+i(sinalpha +sin beta + singamma)` ` = 0+ ixx 0 =0` (a) Now, `(1)/(z_(1)) = (cos alpha + isin alpha)^(-1) = cos alpha- isin alpha` `(1)/(z_(1)) =cos beta- isin beta` `(1)/(z_(2)) =cos gamma-isingamma ` `therefore(1)/(z_(1))+(1)/(z_(2))+(1)/(z_(3))` `=(cos alpha + cos beta + cos gamma) -i(sin alpha + sin beta + sin gamma) (2) ` `=0-ixx 0 =0` `z_(1)^(2) + z_(2)^(2) + z_(3)^(2) =(z_(1) + z_(2) +z_(3))^(2) -2(z_(1)z_(2)+z_(2)z_(3)+z_(3)z_(1))` `=0-2z_(1)z_(2)z_(3)((1)/(z_(3))+(1)/(z_(1)) +(1)/(z_(2)))` `RARR (cos alpha + isin alpha)^(2) + (cos beta + isin beta)^(2) + (cosgamma+isin gamma)^(2) =0` `rArr (cos 2alpha + isin 2alpha)+(cos 2beta+isin2)+(cos 2gamma +isin 2gamma)=0+ixx0` Equating realand imaginary parts on both sides, `cos 2alpha + cos 2beta + cos2gamma =0` `and sin 2alpha + sin 2beta + sin2gamma = 0` (b) `z_(1)^(3) +z_(2)^(3) +z_(3)^(3) =(z_(1) +z_(2))^(3) -3z_(1)z_(2)(z_(1) +z_(2))+z_(3)^(3)` ` = (-z_(3))^(3) -3z_(1)z_(2)(-z_(3))+z_(3)^(3)""["Using (1)"]` `=3z_(1)z_(2)z_(3)` `rArr (cos alpha+sin alpha)^(3)+(cos beta+ isinbeta)^(3) +(cos gamma + isin gamma)^(3)` `= 3(cos alpha + isin alpha) (cos beta+isin beta)(cos gamma + isin gamma)^(3)` `cos 3alpha +isin 3alpha +cos 3beta +isin 3beta + cos3gamma+ isin 3gamma` `= 3{cos (alpha +beta+ gamma)+ isin (alpha + beta+gamma)}` Equaiting imaginary parts on bothsides, `sin 3alpha +sin 3 beta +sin 3gamma =3SIN(alpha + beta + gamma)` (c) Equating real parts on both sides, `cos 3alpha + cos3beta + cos 3gamma = 3cos (alpha+beta+gamma)`

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