Question

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

Answer 1

Let `z_(1) = cos alpha + isin alpha,z_(2) = cos beta + isin beta`,
`z_(3) = cos gamma + isin gamma`
`therefore z_(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 -isin gamma `
`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) + (cos gamma +isin gamma)^(2) =0`
`rArr (cos 2alpha + isin 2alpha)+(cos 2beta+isin2)+(cos 2gamma +isin 2gamma)=0+ixx0`
Equating real and 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+ isin beta)^(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 both sides,
`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)`