Yet another trig question (1 Viewer)

yellowhighlighterr · Jul 11, 2023

I've nailed part a but how do I start part b???
Cheers,
yellowhighlighterr

jimmysmith560 · Jul 11, 2023

Would the following working help?

$\cos3\theta=4\cos^3\theta-3\cos\theta \\ \\ \text{Let} \ x=\cos\theta \\ \\ \cos\frac{2\pi}{3}=-\frac{1}{2} \\ \\ -\frac{1}{2}=4x^3-3x \\ \\ 8x^3-6x+1=0 \\ \\ \cos3\theta=-\frac{1}{2} \\ \\ 3\theta=\frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3} \\ \\ \theta=\frac{2\pi}{9}, \frac{4\pi}{9}, \frac{8\pi}{9} \\ \\ \text{So} \cos\frac{2\pi}{9} \ \text{is a root of} \ 8x^3-6x+1=0.$

yellowhighlighterr · Jul 11, 2023

yesssss dude thank you so much!

Yet another trig question (1 Viewer)

yellowhighlighterr

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jimmysmith560

Le Phénix Trilingue

yellowhighlighterr

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