The Buffling Cosines

Can someone please help me answering this problem. This problem came from the IMO (i think its 1962). Here it goes:

Prove that cosπ7cos2π7+cos3π7=12\cos \frac{\pi}{7} - \cos \frac{2\pi}{7} + \cos \frac{3\pi}{7} = \frac{1}{2}

I have done everything i know with this, even using the sum to product identities, still it doesn't work. Please help. thanks.

Note by Jaydee Lucero
6 years, 2 months ago

No vote yet
12 votes

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

If I recall correctly, the official solution is to consider the expression sinπ7(cosπ7cos2π7+cos3π7), \sin \frac{\pi}{7} \left( \cos \frac{\pi}{7} - \cos \frac{2\pi}{7} + \cos \frac{3\pi}{7} \right), distributing the sine factor and using the product-to-sum identity sinαcosβ=12(sin(α+β)+sin(αβ)). \sin \alpha \cos \beta = \frac{1}{2} \left( \sin (\alpha+\beta) + \sin (\alpha - \beta) \right).

hero p. - 6 years, 2 months ago

Log in to reply

wow!!! :D now i know how will I apply the product to sum identity. But, how should I remove the sine factor thereafter?

Jaydee Lucero - 6 years, 2 months ago

Log in to reply

After you compute and simplify the expression, you will find it equals 12sinπ7 \frac{1}{2} \sin \frac{\pi}{7} , from which the original identity immediately follows.

hero p. - 6 years, 2 months ago

Log in to reply

@Hero P. and now I got it!!! haha :D thanks , men...

Jaydee Lucero - 6 years, 2 months ago

Log in to reply

Since cosπ7=cos6π7\cos\frac\pi 7 = -\cos\frac{6\pi}7 and cos3π7=cos4π7\cos\frac{3\pi}7 = -\cos\frac{4\pi}7, it suffices to prove cos2π7+cos4π7+cos6π7=12\cos\frac{2\pi}7+\cos\frac{4\pi}7+\cos\frac{6\pi}7=-\frac 12

But, 0=1+cos2π7+cos4π7++cos12π7=1+2(cos2π7+cos4π7+cos6π7)0=1+\cos\frac{2\pi}7+\cos\frac{4\pi}7+\dots+\cos\frac{12\pi}7=1+2\left(\cos\frac{2\pi}7+\cos\frac{4\pi}7+\cos\frac{6\pi}7\right) because 1,cos2π7,cos4π7,,cos12π71,\cos\frac{2\pi}7,\cos\frac{4\pi}7,\dots,\cos\frac{12\pi}7 are the real parts of the 7th roots of unity. The result follows.

Brice Huang - 6 years, 2 months ago

Log in to reply

Thanks, very nicely explained.

Lokesh Sharma - 6 years, 2 months ago

Log in to reply

Nice application!

Calvin Lin Staff - 6 years, 2 months ago

Log in to reply

Recall that cos(πk)=cosk\cos(\pi-k)=-\cos k. This means

cosπ7cos2π7+cos3π7(cos6π7cos5π7+cos4π7)=1\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}-\left(\cos\frac{6\pi}{7}-\cos\frac{5\pi}{7}+\cos\frac{4\pi}{7}\right)=1

Rearranging, we get

0=1+cosπ7cos2π7+cos3π7cos4π7+cos5π7cos6π70=-1+\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}-\cos\frac{4\pi}{7}+\cos\frac{5\pi}{7}-\cos\frac{6\pi}{7}

Letting u=eiπ7u=e^\frac{i\pi}{7}, this is

0=Re(1+uu2+u3u4+u5u6)0=\mathrm{Re}(-1+u-u^2+u^3-u^4+u^5-u^6)

But, since u71=(u1)(u6u5+u4u3+u2u+1)=0u^7-1=(u-1)(u^6-u^5+u^4-u^3+u^2-u+1)=0, we know either u=1u=1 or u6u5+u4u3+u2u+1=0u^6-u^5+u^4-u^3+u^2-u+1=0. Clearly u1u\neq1 since Re(u)=sinπ70\mathrm{Re}(u)=\sin\frac{\pi}{7}\neq0, so

u6u5+u4u3+u2u+1=0u^6-u^5+u^4-u^3+u^2-u+1=0

which completes the proof since Re(0)=0\mathrm{Re}(0)=0 as desired.

Cody Johnson - 6 years, 2 months ago

Log in to reply

Use the 7th root of unity...I'm sure you will get the answer..

Rushi Rokad - 6 years, 2 months ago

Log in to reply

How? The cube root of unity would involve i i but there is no ii in the equation to be proved.

Lokesh Sharma - 6 years, 2 months ago

Log in to reply

Recall that cosz=eiz+eiz2. \cos z = \frac{e^{iz} + e^{-iz}}{2}. Also, observe that cos2π7=cos5π7. -\cos \frac{2\pi}{7} = \cos \frac{5\pi}{7}. Now let ζn=e2iπ/n \zeta_n = e^{2i\pi/n} be a primitive nth n^{\rm th} root of unity. Thus cosπ7cos2π7+cos3π7=cosπ7+cos3π7+cos5π7=12(ζ14+ζ141+ζ143+ζ143+ζ145+ζ145)=12(ζ14+ζ1413+ζ143+ζ1411+ζ145+ζ149)=12(ζ147+k=06ζ142k+1)=12(1+ζ14k=06ζ7k)=12(1+ζ140)=12. \begin{aligned} \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} \\ &= \frac{1}{2}(\zeta_{14} + \zeta_{14}^{-1} + \zeta_{14}^3 + \zeta_{14}^{-3} + \zeta_{14}^5 + \zeta_{14}^{-5} ) \\ &= \frac{1}{2}(\zeta_{14} + \zeta_{14}^{13} + \zeta_{14}^3 + \zeta_{14}^{11} + \zeta_{14}^5 + \zeta_{14}^{9} ) \\ &= \frac{1}{2}\left( -\zeta_{14}^7 + \sum_{k=0}^6 \zeta_{14}^{2k+1} \right) \\ &= \frac{1}{2}\left( 1 + \zeta_{14} \sum_{k=0}^6 \zeta_7^k \right) \\ &= \frac{1}{2}(1 + \zeta_{14} \cdot 0) = \frac{1}{2}. \end{aligned} Note that we used the following facts: ζ2n2k=ζnk, \zeta_{2n}^{2k} = \zeta_n^k, and because {1,ζn,ζn2,,ζnn1} \{ 1, \zeta_n, \zeta_n^2, \ldots, \zeta_n^{n-1} \} are the roots of zn1=0 z^n - 1 = 0 , the sum of these roots is the negative of the coefficient of the degree n1 n-1 term, which is zero for any positive integer n2 n \ge 2 .

I think that you should be able to tell which solution method I prefer.

hero p. - 6 years, 2 months ago

Log in to reply

ill try this one... :D

Jaydee Lucero - 6 years, 2 months ago

Log in to reply

cos2π7+cos4π7+cos6π7\cos \frac{ 2\pi }{ 7 }+\cos \frac{ 4\pi }{ 7 }+\cos \frac{ 6\pi }{ 7 }sin3π7cos3π7sinπ7\frac{ \sin \frac{ 3\pi }{ 7 }*\cos \frac{ 3\pi }{ 7 } }{ \sin \frac{ \pi }{ 7 } }sin6π72sinπ7\frac{ \sin \frac{ 6\pi }{ 7 } }{ 2\sin \frac{ \pi }{ 7 } }12\frac{ 1 }{ 2 }

Cody Martin - 6 years, 1 month ago

Log in to reply

right

Sonu Kumar Tiwari - 6 years, 1 month ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...