Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN: 9781133382119
Author: Swokowski
Publisher: Cengage
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[画像: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.]
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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.
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