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December 17th, 2018, 01:54 PM   #1
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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
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December 17th, 2018, 03:44 PM   #2
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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.
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December 17th, 2018, 08:05 PM   #3
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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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