# Adv Alegbra HW Solutions 145 - 145 Solution We know that...

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145 Solution. We know that cos ϕ is a root of f ( x ) = 4 x 3 3 x r , where r = cos 3 ϕ . In particular, if ϕ = θ + 120 , then cos ϕ is a root of 4 x 3 3 x cos 3 + 120 ) . But the addition formula for cosine gives cos 3 + 120 ) = cos 3 θ = r , and so cos + 120 ) is also a root of f ( x ) . Similarly, cos 3 + 240 ) = cos 3 θ , and cos + 240 ) is also a root of f ( x ) . 5.9 (i) Prove that cosh ( 3 θ) = 4 cosh 3 (θ) 3 cosh (θ) . Solution. By de f nition, cosh θ = 1 2 ( e θ + e θ ) . Expand and then simplify 4 [ 1 2 ( e θ + e θ ) ] 3 3 [ 1 2 ( e θ + e θ ) ] to obtain 1 2 ( e 3 θ + e 3 θ ) . (ii) Prove that sinh ( 3 θ) = 4 sinh 3 (θ) + 3 sinh (θ) . Solution. Absent. 5.10 Find the roots of x 3 9 x + 28. Solution. The cubic formula works smoothly: q =− 9, r = 28, D = 676, α =− 1, β = 3, and the roots are 4 and 2 ± 3. 5.11 Find the roots of x 3 24 x 2 24 x 25. Solution. After substituting X = x + 8, one obtains x 3 216 x 1241. For the new cubic, q =− 216, r =− 1241, D = 47089 = ( 217 ) 2 , α = 9, β =− ( 216 )/ 3 · 8, and so a root is 17; it follows that a root of the original cubic is 17 + 8 = 25. The division algorithm gives X 3 24 X 2
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## This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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