Show $4 \cos^2{\frac{\pi}{5}} - 2 \cos{\frac{\pi}{5}} -1 = 0$

Show

$$4 \cos^2{\frac{\pi}{5}} - 2 \cos{\frac{\pi}{5}} -1 = 0$$

The hint says "note $\sin{\frac{3\pi}{5}} = \sin{\frac{2\pi}{5}}$" and "use double/triple angle or otherwise"

So I have

$$4 \cos^2{\frac{\pi}{5}} - 2 (2 \cos^2{\frac{\pi}{10}} - 1) - 1$$

$$4 \cos^2{\frac{\pi}{5}} - 4 \cos^2{\frac{\pi}{10}} +1$$

Now what? Theres $\frac{\pi}{5}$ and $\frac{\pi}{10}$ and I havent used the tip on $\sin$ so perhaps I am missing something?

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The hint says to use the following:

$$\sin 3x = 3 \sin x - 4 \sin^3x$$

and

$$\sin 2x = 2 \sin x \cos x$$

So if $x = \frac{\pi}{5}$, then we have, using $\sin 3x = \sin 2x$ that

$$3 \sin x - 4 \sin^3 x = 2 \sin x \cos x$$

Since $\sin x \neq 0$, cancel it, and use $\sin^2 x = 1 - \cos ^2x$.

For a geometric way to find $\cos \frac{\pi}{5}$ see: http://www.cut-the-knot.org/pythagoras/cos36.shtml

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Here is an alternate approach.

Using de Moivre's Formula, we get for $\theta=\frac{\pi}{5}$ $$0=(\cos(\theta)+i\sin(\theta))^5+1\tag{1}$$ Looking at the real part of $(1)$ yields \begin{align} 0 &=\color{red}{\cos^5(\theta)}\color{green}{-10\cos^3(\theta)\sin^2(\theta)}\color{blue}{+5\cos(\theta)\sin^4(\theta)}+1\\ &=\color{red}{\cos^5(\theta)}\color{green}{-10\cos^3(\theta)+10\cos^5(\theta)}\color{blue}{+5\cos(\theta)-10\cos^3(\theta)+5\cos^5(\theta)}+1\\ &=16\cos^5(\theta)-20\cos^3(\theta)+5\cos(\theta)+1\\ &=(4\cos^2(\theta)-2\cos(\theta)-1)^2(\cos(\theta)+1)\tag{2} \end{align} Since $\cos(\theta)=-1$ only when $\theta$ is an odd multiple of $\pi$, we are left with $$4\cos^2\left(\frac{\pi}{5}\right)-2\cos\left(\frac{\pi}{5}\right)-1=0\tag{3}$$

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Use the identities $$\sin 2\theta=2\cos\theta\sin\theta$$

and \eqalign{\sin 3\theta &=\sin(2\theta+\theta)\cr &= \sin2\theta \cos\theta+\sin\theta\cos2\theta\cr &= 2\sin\theta\cos^2\theta +\sin\theta(2\cos^2\theta-1)\cr } to write the identity $$\textstyle\sin{2\pi\over5}=\sin(\pi-{2\pi\over5})=\sin{3\pi\over5}$$ as $$\tag{1}\textstyle 2\sin{\pi\over5}\cos{\pi\over5} = 2\sin{\pi\over5}\cos^2{\pi\over5} +\sin{\pi\over5}(2\cos^2{\pi\over5}-1).$$ Cancelling the term $\sin{\pi\over5}\ne0$ from both sides of equation $(1)$ gives: $$\textstyle 2 \cos{\pi\over5} = 2\cos^2{\pi\over5} + (2\cos^2{\pi\over5}-1);$$ or, $$\textstyle 0 = 4\cos^2{\pi\over5} -2\cos{\pi\over5} -1 ,$$ as desired.

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You don't have to use the hint. Let $x = {\pi \over 5}$. Then $\cos(3x) = -\cos(2x)$ since $3x + 2x = \pi$. Since $\cos(3x) = 4\cos^3(x) - 3\cos(x)$ and $\cos(2x) = 2\cos^2(x) - 1$, you get $$4\cos^3(x) - 3\cos(x) = - ( 2\cos^2(x) - 1)$$ This simplifies to $$4\cos^3(x) + 2\cos^2(x) - 3\cos(x) - 1 = 0$$ This factors into $$(4\cos^2(x) - 2\cos(x) - 1)(\cos(x) + 1) = 0$$ Since $\cos(x)$ is not $-1$, you get the desired equation $4\cos^2(x) - 2\cos(x) - 1 = 0$.

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