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Prove that; $8\cos^3 (\frac {\pi}{9})- 6\cos(\frac {\pi}{9})=1$

My Attempt,

$$L.H.S=8\cos^3(\frac {\pi}{9}) - 6\cos(\frac {\pi}{9})$$ $$=2\cos(\frac {\pi}{9}) [4\cos^2(\frac {\pi}{9}) - 3]$$ $$=2\cos(\frac {\pi}{9}) [2+2\cos(\frac {2\pi}{9}) - 3]$$ $$=2\cos(\frac {\pi}{9}) [2\cos(\frac {2\pi}{9})-1]$$.

What should I do further?

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    $\begingroup$ Hint: $\cos 3x=4\cos^{3}x-3\cos x$. $\endgroup$ – Paramanand Singh Jan 29 '17 at 8:33
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$2(4 \cos^3 \frac π9 - 3 \cos \frac π9)$

As $\cos 3\theta = 4 \cos^3 \theta - 3\cos \theta$

= $2(\cos (3 × \frac π9))$

=$2 × \cos(\frac π3)$

=$2 × \frac 12 = 1$

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Hint:

$$\cos(3x) = 4\cos^3(x) - 3\cos(x).$$

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$\cos 3x=4\cos^3x-3\cos x \Rightarrow 2\cos 3x=8\cos^3x-6\cos x$.

You can now see that if you put $x=\frac{\pi}{9}$, you get the result.

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\begin{align} 8\cos^3 (\frac {\pi}{9})- 6\cos(\frac {\pi}{9}) &=8\cos(\frac{\pi}9)\left[1-\sin^2(\frac{\pi}9)\right]-6\cos^(\frac{\pi}9)\\ &=2\cos(\frac{\pi}9)-8\cos(\frac{\pi}9)\sin^2(\frac{\pi}9)\\ &=2\cos(\frac{\pi}9)-4\sin(\frac{2\pi}9)\sin(\frac{\pi}9)\tag1\\ &=2\cos(\frac{\pi}9)-2\left[\cos(\frac{\pi}9)-\cos(\frac{3\pi}9)\right]\tag2\\ &=2\cos(\frac{\pi}3)=1 \end{align}

Note:

$(1)$:(Double Angle Formula) $\sin{2t}=2\sin t\cos t$

$(2)$:(Product-to-sum-formula) $2\sin a\sin b=\cos(a-b)-\cos(a+b)$

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