How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ without using a calculator.
Related question: how do we prove that $\cos(\pi/5)\cos(2\pi/5) = 0.25$, also without using a calculator
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The complex roots of $x^5-1$ are: $$ \begin{align} x_1&=1\\ x_2&=\cos\frac{2\pi}5+i\sin\frac{2\pi}5\\ x_3&=-\cos\frac{\pi}5+i\sin\frac{\pi}5\\ x_4&=-\cos\frac{\pi}5-i\sin\frac{\pi}5\\ x_5&=\cos\frac{2\pi}5-i\sin\frac{2\pi}5 \end{align} $$ using Vieta's formulas you get $$0=x_1+x_2+\dots+x_5=1+2\left(\cos\frac{2\pi}5-\cos\frac\pi5\right)=0,$$ which yields your first equation. From now on, let $\varphi=\frac{\pi}5$ (for brevity). We know that $\cos2\varphi-\cos\varphi+\frac12=(2\cos^2\varphi-1)-\cos\varphi+\frac12=2\cos^2\varphi-\cos\varphi-\frac12=0$, i.e. $$2\cos^2\varphi-\cos\varphi=\frac12.$$ Now from $\cos2\varphi=\cos\varphi-\frac12$ you get $$\cos\varphi\cos2\varphi=\cos^2\varphi-\frac{\cos\varphi}2=\frac{2\cos^2\varphi-\cos\varphi}2=\frac14.$$ (Or, as suggested in Chandrasekhar's answer, from $2\cos^2\varphi-\cos\varphi=\frac12$ you can find the value of $\cos\varphi$ by solving the quadratic equation and taking the positive root. Once you know $\cos\varphi$, you can compute $\cos2\varphi$ and many other things. If you try it this way, you can check your result e.g. here.) |
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Note that: $\cos{2x} = \cos^{2}{x} - \sin^{2}{x} = 2\:\cos^{2}{x} - 1$. Therefore you have $\cos \frac{2\pi}{5} = 2\:\cos^{2}\frac{\pi}{5} - 1$ Now, \begin{align*} \cos\frac{\pi}{5} - \cos\frac{2\pi}{5} = \cos\frac{\pi}{5} - 2\: \cos^{2}\frac{\pi}{5}+1 \end{align*} This is a quadratic equation of the form $2 x^{2} - x -1 =0$ and solving this will give you the value of $\cos\frac{\pi}{5}$ from which you can find the above value which you need. |
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If $a=\cos\frac{\pi}5$ and $b=\cos\frac{2\pi}5=2a^2-1$ (by the double-angle identity for cosine), we want to show that $$ 0=(x-a)(x+b)=x^2-(a-b)x-ab=x^2-\tfrac12x-\tfrac14, $$ i.e., that $a$ and $-b$ are roots of $4x^2-2x-1$. Note also that $4x^2+2x-1=4(x+a)(x-b)$ has roots $-a$ and $b$. What is special about the numbers $\{a,b,-b,-a\}$? They are the $x$-coordinates (real parts) of the nonreal $10$th roots of unity, $$ x+iy=e^{\pm\pi i\cdot\frac{k}5} \qquad \text{for} \qquad k\not\equiv0\pmod5. $$ But these satisfy the equation $(x+iy)^5=(-1)^k=\pm1$. Taking the imaginary part, we have $$ \eqalign{ 0 &= \Im\left[(x+iy)^5\right] \\ &= \Im\left[x^5+5x^4(iy)+10x^3(iy)^2+10x^2(iy)^3+5x(iy)^4+(iy)^5\right] \\ &= \Im\left[iy\left(5x^4+10x^2(iy)^2+(iy)^4\right)\right] \\ &= y\left[5x^4-10x^2y^2+y^4\right] \\ \implies 0 &= \left[5x^4-10x^2(1-x^2)+(1-x^2)^2\right] \\ &= 16x^4-12x^2+1 \\ &= 16x^4-8x^2+1 ~-~4x^2 \\ &= \left( 4x^2-1 \right) - \left( 2x \right)^2 \\ &= \left( 4x^2+2x-1\right)\left( 4x^2-2x-1\right) \,. } $$ So far, we have shown this has roots $\{\pm a,\pm b\}$. It only remains to show that $\{-a,b\}$ are the roots of the first factor and that the desired pair $\{a,-b\}$ splits the second quadratic factor into linear terms (as we wrote at the outset). Perhaps there is a clever way to infer the correct linear order of this set of roots by noticing that $a>\cos\frac\pi4>b$ so that $a^2>\frac12>b^2$ and combining this with our identity $b=2a^2-1$ above. I propose in stead to use that $a>b$ but then to notice that our quadratic factors are in fact parabolas, with roots that are very easy to order on the $x$-axis. If we let $t=2x$ (which preserves order), then we have $$ t^2\pm t-1 = t\,(t\pm1)-1 $$ which both pass through $(0,-1)$ and alternately pass through $(\mp1,-1)$, shown respectively in red and blue below; the result follows.
It's only a small extra step to note that the roots of our factor, $t^2-t-1$, are $t=\frac{1\pm\sqrt5}{2}$, or $x=\frac{1\pm\sqrt5}{4}$. |
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For the first equality. |
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$$\cos(\pi/5) - \cos(2\pi/5)= {\rm Re}(e^{i\frac{\pi}{5}}-e^{i\frac{2\pi}{5}})=\frac{1}{2}( e^{i\frac{\pi}{5}}-e^{i\frac{2\pi}{5}}+ \overline{e^{i\frac{\pi}{5}}-e^{i\frac{2\pi}{5}}})=\frac{1}{2}( e^{i\frac{\pi}{5}}-e^{i\frac{2\pi}{5}}+ e^{-i\frac{\pi}{5}}-e^{-i\frac{2\pi}{5}})=\frac{e^{\frac{-2i\pi}{5}}}{2}( -e^{i\frac{4\pi}{5}}+e^{i\frac{3\pi}{5}}+ e^{i\frac{\pi}{5}}-1)$$ To simplify the computations, let $\omega=e^{\frac{i \pi}{5}}$. Note that $\omega^5=-1$. Then $$\cos(\pi/5) - \cos(2\pi/5)= \frac{-1}{2\omega^2}(\omega^4-\omega^3-\omega+1)= \frac{-1}{2\omega^2}(\omega^4-\omega^3+\omega^2-\omega+1-\omega^2)$$ $$\cos(\pi/5) - \cos(2\pi/5)= \frac{-1}{2\omega^2}(\frac{\omega^5+1}{\omega+1} -\omega^2)=\frac{-1}{2\omega^2}(0 -\omega^2)=\frac{1}{2}$$ |
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I was not able (yet) to follow Chandrasekar's solution, but noticed this while trying to understand the argument (how it could possibly lead to the solution, or how exactly he arrives at $2x^2-x-1$ for $x=\cos\frac{\pi}{5}$, which to me seems non-obvious and even a fallacious deduction from his equations and prose -- apologies if I am just being dense)...perhaps it is what Chandrasekar meant all along, but in any case, it does seem to be the most elementary solution available. Apply the double angle formula $\cos2\theta=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1$ to $\theta=\frac{\pi}{5}$ and $\frac{2\pi}{5}$, with $a=\cos\frac{\pi}{5}$ and $b=\frac{2\pi}{5}$ for convenience, recalling also that $\cos(\pi\pm\theta)=-\cos\theta$: $$ b=\cos\frac{2\pi}{5}=2\,\cos^2\frac{\pi}{5}-1=2a^2-1 $$ $$ -a=\cos\frac{4\pi}{5}=2\,\cos^2\frac{2\pi}{5}-1=2b^2-1 $$ Next, subtracting the equations $$ \matrix{ 2a^2=1+b\\ 2b^2=1-a} $$ we get $$ \eqalign{ 2\left(a^2-b^2\right)&=b+a\\ 2\left(a+b\right)\left(a-b\right)&=b+a\\ 2\left(a-b\right)&=1\\ a-b&=\frac12\,. } $$ Furthermore, multiplying, we get $$ 4(ab)^2=(1+b)(1-a)=1+(b-a)-ab=1+\left(-\tfrac12\right)-ab $$ giving us the quadratic equation $$4(ab)^2+(ab)-\tfrac12=0$$ $$8(ab)^2+2(ab)-1=0$$ $$\left(4ab-1\right)\left(2ab+1\right)=0$$ so that $ab=\frac14$ or $-\frac12$, from which we can choose the former since we know that $0<a<b$. |
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$$\cos(\pi/5)\cos(2\pi/5)=A$$ $$\Longrightarrow\quad A = \frac{\sin(\pi/5)\cos(\pi/5)\cos(2\pi/5)}{\sin(\pi/5)}=\frac{\sin(2\pi/5)\cos(2\pi/5)}{2\sin(\pi/5)}$$ $$A = \frac{\sin(4\pi/5)} {2\cdot 2\cdot\sin(\pi/5)}=\frac{1}{4}$$ since $\sin(4\pi/5)=\sin(\pi/5)$. |
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