Problem 1.A (Application of de Moivre's Theorem: Finding the roots of a cubic polynomial) Notice that for any cubic polynomial dividing by a and substituting x = y — 3b а ax3 + bx²+cx+d, a 0, gives a cubic polynomial in depressed form, y³ + py + q. For this depressed cubic, the discriminant (the analogue of the quadratic discriminant b² - 4ac of ax² + bx + c) is _ A = -4 -4p³ - 27q², and as with the quadratic discriminant the roots of the quadratic are all real and distinct if A > 0. Henceforth assume that A >0. (An analogous technique can be used to handle the case A < 0.) a. Expand e30 in two different ways to establish the identity. cos 30 = 4 cos³ 0 - 3 cos 0. Remark: The same method yields general formulae for cos no and sinne, nЄZ, as polynomials in sin 0, cos 0. b. We solve the depressed cubic for y by exploiting the identity in (a); with solutions y in hand we can recover x by reversing the above change x y of variables. (i) Substitute y = A cos in our depressed cubic. (ii) Rearrange the equation and find a value for A (in terms of p) so that the equation becomes 4 cos³ 0-3 cos 0 = K for some constant K written in terms of p, q. (iii) Using the identity from (a) our equation becomes cos 30 = K. Find all 3 solutions in terms of p, q. (iv) Write the solutions in terms of y. c. The discriminant of the polynomial ƒ(y) := y³ − 3y+1 (already in depressed form) is A = 81 > 0. Use the technique in (b) to express the roots of the polynomial in terms of cosines of rational multiples of 7.
Problem 1.A (Application of de Moivre's Theorem: Finding the roots of a cubic polynomial) Notice that for any cubic polynomial dividing by a and substituting x = y — 3b а ax3 + bx²+cx+d, a 0, gives a cubic polynomial in depressed form, y³ + py + q. For this depressed cubic, the discriminant (the analogue of the quadratic discriminant b² - 4ac of ax² + bx + c) is _ A = -4 -4p³ - 27q², and as with the quadratic discriminant the roots of the quadratic are all real and distinct if A > 0. Henceforth assume that A >0. (An analogous technique can be used to handle the case A < 0.) a. Expand e30 in two different ways to establish the identity. cos 30 = 4 cos³ 0 - 3 cos 0. Remark: The same method yields general formulae for cos no and sinne, nЄZ, as polynomials in sin 0, cos 0. b. We solve the depressed cubic for y by exploiting the identity in (a); with solutions y in hand we can recover x by reversing the above change x y of variables. (i) Substitute y = A cos in our depressed cubic. (ii) Rearrange the equation and find a value for A (in terms of p) so that the equation becomes 4 cos³ 0-3 cos 0 = K for some constant K written in terms of p, q. (iii) Using the identity from (a) our equation becomes cos 30 = K. Find all 3 solutions in terms of p, q. (iv) Write the solutions in terms of y. c. The discriminant of the polynomial ƒ(y) := y³ − 3y+1 (already in depressed form) is A = 81 > 0. Use the technique in (b) to express the roots of the polynomial in terms of cosines of rational multiples of 7.
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section: Chapter Questions
Problem 7DE
Related questions
Question
No chatgpt pls will upvote
Transcribed Image Text:Problem 1.A (Application of de Moivre's Theorem: Finding the roots of a cubic polynomial) Notice
that for any cubic polynomial
dividing by a and substituting x = y —
3b
а
ax3 + bx²+cx+d,
a 0,
gives a cubic polynomial in depressed form,
y³ + py + q.
For this depressed cubic, the discriminant (the analogue of the quadratic discriminant b² - 4ac of ax² + bx + c) is
_
A = -4
-4p³ - 27q²,
and as with the quadratic discriminant the roots of the quadratic are all real and distinct if A > 0. Henceforth
assume that A >0. (An analogous technique can be used to handle the case A < 0.)
a. Expand e30 in two different ways to establish the identity.
cos 30 = 4 cos³ 0 - 3 cos 0.
Remark: The same method yields general formulae for cos no and sinne, nЄZ, as polynomials in sin 0, cos 0.
b. We solve the depressed cubic for y by exploiting the identity in (a); with solutions y in hand we can recover
x by reversing the above change x y of variables.
(i) Substitute y
=
A cos in our depressed cubic.
(ii) Rearrange the equation and find a value for A (in terms of p) so that the equation becomes
4 cos³ 0-3 cos 0 = K
for some constant K written in terms of p, q.
(iii) Using the identity from (a) our equation becomes cos 30 = K. Find all 3 solutions in terms of p, q.
(iv) Write the solutions in terms of y.
c. The discriminant of the polynomial ƒ(y) := y³ − 3y+1 (already in depressed form) is A = 81 > 0. Use the
technique in (b) to express the roots of the polynomial in terms of cosines of rational multiples of 7.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 6 images
Recommended textbooks for you
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage
Algebra: Structure And Method, Book 1
Algebra
ISBN:
9780395977224
Author:
Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:
McDougal Littell
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage
Algebra: Structure And Method, Book 1
Algebra
ISBN:
9780395977224
Author:
Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:
McDougal Littell
College Algebra (MindTap Course List)
Algebra
ISBN:
9781305652231
Author:
R. David Gustafson, Jeff Hughes
Publisher:
Cengage Learning
Algebra and Trigonometry (MindTap Course List)
Algebra
ISBN:
9781305071742
Author:
James Stewart, Lothar Redlin, Saleem Watson
Publisher:
Cengage Learning