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 December 17th, 2018, 12:54 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 07:07 PM. Reason: to change 8 to -8
 December 17th, 2018, 02:44 PM #2 Math Team     Joined: Jul 2011 From: Texas Posts: 3,016 Thanks: 1600 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, 07:05 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,968 Thanks: 2217 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.

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