i think the sum of these three cosin is equal to 1/2.

dashmiz wrote: i think the sum of these three cosin is equal to 1/2. the question is to prove it

$\cos x - \cos 2x + \cos 3x = \frac{1}{2}$ $\cos x + \cos 3x + \cos 5x = \frac{1}{2}$ $\cos x + \cos 3x + \cos 5x + ... = \frac{{\sin 2nx}}{{2\sin x}}$ $n = 3$ $\cos x + \cos 3x + \cos 5x = \frac{{\sin 6x}}{{2\sin x}}$ $\frac{{\sin 6x}}{{2\sin x}} = \frac{1}{2} \Leftrightarrow \sin 6x = \sin x$ $x = \frac{{2k\pi }}{5}(1)\quad x = \frac{{(2k + 1)\pi }}{7}(2)$ $(2)\;k = 0 \to x = \frac{\pi }{7}$ $.^..$ $\cos \frac{\pi }{7} - \cos \frac{{2\pi }}{7} + \cos \frac{{3\pi }}{7} = \frac{1}{2}$

Does anyone know how to find the polynomial in which those values are the roots?

Sorry for reviving this very old topic but I was looking through old problems and saw pkothari13's question. What I did was to consider the roots of unity of $ z^{7}=1$ so $ z= cis \frac{2 \pi k}{7}$ where$ k=-3,-2,-1,0,1,2,3$. Now consider the sum of roots: $ 0 = 1+2( \cos{\frac{2 \pi}{7}}+\cos{\frac{4 \pi}{7}}+\cos{\frac{6 \pi}{7}})$ Now since: $ \cos{\frac{6 \pi}{7}}=-\cos{\frac{\pi}{7}}$ and $ \cos{\frac{4 \pi}{7}}=-\cos{\frac{3 \pi}{7}}$ We have $ \cos{\frac{\pi}{7}}-\cos{\frac{2 \pi}{7}}+\cos{\frac{3 \pi}{7}}= \frac{1}{2}$.

DPopov wrote: Prove that $ \cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=\frac{1}{2}$ Let : $ S_{1}=LS=cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=cos{\frac{\pi}{7}}+\cos{\frac{3\pi}{7}}+\cos{\frac{5\pi}{7}}$ Then we have: $ 2sin{\frac{\pi}{7}}S_{1}=2sin{\frac{\pi}{7}}(cos{\frac{\pi}{7}}+\cos{\frac{3\pi}{7}}+\cos{\frac{5\pi}{7}})$ Use : $ 2sinA.cosB=sin(A+B)+sin(A-B)$ We have: $ 2sin{\frac{\pi}{7}}S_{1}=sin{\frac{2\pi}{7}+sin{\frac{4\pi}{7}-sin{\frac{2\pi}{7}+sin{\frac{6\pi}{7}-sin{\frac{4\pi}{7}=sin{\frac{6\pi}{7}=sin{\frac{\pi}{7}}}}}}}}$ So : $ S_{1}=\frac{1}{2}$ And we are done!!

On Oct 30th, I took the 1st grade of Japanese Suugaku Kentei (The Mathematics Certification). http://www.su-gaku.net/ In Question [2] of the first exam, I encountered the similar problem. Find the polynomial $P(x)$ of the lowest degree such that (1) the coefficient of the highest degree is 1 (2) all coefficients are rational (3)$P \left( 2 \cos \frac{2\pi}{7} \right)=0$

The $\text{LHS}$ of the desired identity is $K=\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7}$. Now observe that \[K\sin\frac{\pi}{7}=\frac{\sin\frac{2\pi}{7}}{2}+\frac{\sin\frac{4\pi}{7}-\sin\frac{2\pi}{7}}{2}+\frac{\sin\frac{6\pi}{7}-\sin\frac{4\pi}{7}}{2}=\frac{\sin\frac{6\pi}{7}}{2}\implies K=\frac{1}{2}\] as desired. $\Box$

1997,I proof: $\cos^k \frac{\pi }{7} + \cos^k \frac{{3\pi }}{7} + \cos^k \frac{{5\pi }}{7} = \frac{1}{2}.$(k=1,3,5) Let $n$ be positive integer ,$ k=1,3,5,...,2n-1 $, prove that $\cos^k \frac{\pi }{2n+1} + \cos^k \frac{{3\pi }}{2n+1}+... + \cos ^k \frac{{(2n-1)\pi }}{2n+1} = \frac{1}{2} .$

tc1729 wrote: The $\text{LHS}$ of the desired identity is $K=\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7}$. Now observe that \[K\sin\frac{\pi}{7}=\frac{\sin\frac{2\pi}{7}}{2}+\frac{\sin\frac{4\pi}{7}-\sin\frac{2\pi}{7}}{2}+\frac{\sin\frac{6\pi}{7}-\sin\frac{4\pi}{7}}{2}=\frac{\sin\frac{6\pi}{7}}{2}\implies K=\frac{1}{2}\]as desired. $\Box$ Could someone explain this to me in more detail? Is it a result of the product sum formulas?

gary2003 wrote: tc1729 wrote: The $\text{LHS}$ of the desired identity is $K=\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7}$. Now observe that \[K\sin\frac{\pi}{7}=\frac{\sin\frac{2\pi}{7}}{2}+\frac{\sin\frac{4\pi}{7}-\sin\frac{2\pi}{7}}{2}+\frac{\sin\frac{6\pi}{7}-\sin\frac{4\pi}{7}}{2}=\frac{\sin\frac{6\pi}{7}}{2}\implies K=\frac{1}{2}\]as desired. $\Box$ Could someone explain this to me in more detail? Is it a result of the product sum formulas? Yes.

Let $z = \cos \frac{\pi}{7} + i \sin \frac{\pi}{7}$. Problem is to prove that the real part of the number $z-z^2+z^3$ equals to $\frac{1}{2}$. Note that $z^7 = -1$ and $z \neq -1$. Introducing $w = z^3-z^2+z$, we have $z^3w-w+1=0$ or \[ w = \frac{1}{1-z^3} \]Then the proof is completed by the following general observation; lemma: For every complex number $\alpha \neq 1$ with absolute value equals to $1$, the real part of $\frac{1}{1-\alpha}$ equals to $\frac{1}{2}$. <Proof> Let $\beta = \frac{1}{1-\alpha}$. It suffices to show that $\beta + \overline{\beta} =1$. Actually \[ \beta + \overline{\beta} = \frac{1-\overline{\alpha} + 1- \alpha}{(1-\alpha)(1-\overline{\alpha})} =1 \]$\square$

Prove that $$ \cos{\frac{\pi}{13}}+\cos{\frac{3\pi}{13}}+\cos{\frac{9\pi}{13}}=\frac{\sqrt{13}+1}{4}.$$$$\cos\frac{2\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}=\frac{\sqrt{13}-1}{4}$$here here here https://artofproblemsolving.com/community/c2231h1379091 https://socratic.org/questions/how-do-you-prove-cos-pi-7-cos-2pi-7-cos-3pi-7-1-8

redacted.

Suppose $w = cis\left(\frac{\pi}{7}\right)$. Then \begin{align*} &\cos \frac{\pi}{7} - \cos \frac{2 \pi}{7} + \cos \frac{3 \pi}{7} \\ &= \frac 12 \left((w+w^{13}) - (w^2 + w^{12}) + (w^3 + w^{11})\right) \\ &= \frac 12 \left(w^{11} + w\right)\left(w^2-w+1\right) \\ &= \frac{(w^3+1)(-w^4+w)}{2(w+1)} \\ &= \frac{w-w^7}{2(w+1)} \\ &= \frac 12.~\blacksquare \end{align*}