AWESOME ===>de moivre's thorem and polynomials<==== AweSOME! (1 Viewer)

[243_robbo]

use de movre's theorem to express cos 4θ in terms of cos θ and use your results to solve the equation 8x^4 - 8x^2 + 1 = 0.

hence show that:

cos π/8 + cos5π/8 + cos 7π/8 = 0

cos π/8 . cos 3π/8 . cos 5π/8 . cos 7π/8 = 1/8

(π is supposed to be pi)

 

[243_robbo]

dont worry i got it

 

[243_robbo]

actually i dont get it i can only get as far as the first step via binomial expansion, but where do the angles come in?
..... riviet?

 

[shsshs]

...........................................4
easy. expand (cosθ + isinθ) using de movire's and binomial


then equate real parts
......................2......................2
turn all the sin θ's into 1 - cos θ
.......................................4..........2
then solve cos4θ = 8cos θ - 8cos θ + 1 = 0

your roots should be cosπ, cos3π cos5π cos7π
.....................................8.........8.........8.........8

using sum of roots you shld get ur answers..

btw i think it shld be hence show that:

cos π/8 + cos3π/8 + cos5π/8 + cos 7π/8 = 0

 

[pLuvia]

And this should be the product of the roots you get

cos π/8 . cos 3π/8 . cos 5π/8 . cos 7π/8 = 1/8

 

[Shady01]

wee wee I remembeer this question FiTzpatrick question 4 i think 36(c) try question 12 thats hard lol

 

[pLuvia]

I remember doing this in the cambridge 4u book but I guess it could be the same