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 December 17th, 2018, 01:54 PM #1 Newbie   Joined: Dec 2018 From: Canada Posts: 14 Thanks: 0 Math Focus: mainly algebra Showing Roots in R P (x) = -8x³ + 6x - √2 Using: sin (3x) = - 4sin³(x) + 3sin (x) -Show that sin (π / 12); sin (17π / 12); sin (3π / 4) are zeros of P in R. -Determine the value of sin (17π / 12) × sin (3π / 4) × sin (π / 12) in two different ways. π = pi Some help, please? Last edited by skipjack; December 17th, 2018 at 08:07 PM. Reason: to change 8 to -8 December 17th, 2018, 03:44 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 3,101 Thanks: 1677 Note ... $P(x) = -8x^3+6x-\sqrt{2}$, the leading coefficient is $-8$ let $x = \sin{t}$ $0 = -8\sin^3{t} + 6\sin{t} - \sqrt{2}$ $0 = -4\sin^3{t} + 3\sin{t} - \dfrac{\sqrt{2}}{2}$ $0 = \sin(3t) - \dfrac{\sqrt{2}}{2} \implies \sin(3t) = \dfrac{\sqrt{2}}{2}$ Note the solution $3t = \dfrac{\pi}{4} + k \cdot 2\pi \, \text{ where } k \in \{0, 1, 2 \}$ will yield the angles given in the problem. Thanks from topsquark December 17th, 2018, 08:05 PM #3 Global Moderator   Joined: Dec 2006 Posts: 21,124 Thanks: 2332 sin(17π/12) × sin(3π/4) × sin (π/12) = product of zeros of P(x) = - √2 / 8. sin(17π/12)sin(π/12)sin(3π/4) = (cos(4π/3) - cos(3π/2))/2 × √2/2 = (-1/2 - 0)/2 × √2/2 = - √2/8. Tags roots, showing Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post annakar Calculus 8 December 16th, 2012 05:52 AM restin84 Algebra 3 March 27th, 2012 11:19 PM happygolucky88 Linear Algebra 1 September 1st, 2009 09:12 PM jamesuminator Algebra 15 December 19th, 2008 09:50 AM

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