What Does Solving Sqrt[7]^2 == r^2 + r^2 - 2 * r * r * Cos[2 * Pi / 3] Involve?

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Discussion Overview

The discussion revolves around solving the equation Sqrt[7]^2 == r^2 + r^2 - 2 * r * r * Cos[2 * Pi / 3], which involves concepts from trigonometry and the law of cosines. Participants explore various approaches to simplify and solve the equation, including drawing triangles and applying the law of sines.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Some participants clarify the equation format, confirming it as {(\sqrt 7)}^2 = r^2 + r^2 - 2rr \cos{(\frac {2 \pi} 3)}.
  • One participant suggests using the law of sines instead of the law of cosines, indicating it may be simpler.
  • Another participant proposes multiplying out the terms and factoring out r^2 to solve the equation.
  • There is a suggestion to simplify the left-hand side (LHS) and collect terms on the right-hand side (RHS) to form a quadratic equation.
  • One participant mentions the identity 1-cos(theta) = 2sin^2(theta/2) as a potential simplification tool.
  • A later reply questions whether the goal is to find r or something related to the triangle, noting that cos(2pi/3) = -1/2 makes the rest of the problem easier.

Areas of Agreement / Disagreement

Participants express varying interpretations of the problem and different methods for approaching the solution. There is no consensus on the best approach or whether the problem is fully understood.

Contextual Notes

Some assumptions about the problem's context and the definitions of terms may be missing. The discussion does not resolve the mathematical steps or the exact nature of the problem being solved.

IndustriaL
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I have no idea how to do this problem can you guys shed some light?

Sqrt[7]^2 == r^2 + r^2 - 2 * r * r * Cos[2 * Pi / 3]

Thanks.
 
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Do you mean:

[tex]{(\sqrt 7)}^2 = r^2 + r^2 -2rr \cos{(\frac {2 \pi} 3)}[/tex]
?
 
well first of all, that's the law of cosines. if i were you, i'd draw your triangle out, and use the law of sines which is simpler.
if you really need to solve it that way, its pretty simple. multiply everything out, and then factor out r^2 and solve.

not really sure where you got stuck, but you also have to show your work here.
 
Is this the problem :

[tex]\sqrt{7}^2 = r^2 + r^2 - 2r^2 cos \left( \frac{2 \pi}{3} \right)[/tex] ?

Can you not simplify the LHS, collect terms in the RHS and plug in the value of the cosine to end up with a trivial quadratic ? Or did I misunderstand the question ?
 
Gokul43201 said:
Is this the problem :

[tex]\sqrt{7}^2 = r^2 + r^2 - 2r^2 cos \left( \frac{2 \pi}{3} \right)[/tex] ?

Can you not simplify the LHS, collect terms in the RHS and plug in the value of the cosine to end up with a trivial quadratic ? Or did I misunderstand the question ?
Nope! :biggrin:

Also, 1-cos(theta) = 2sin^2(theta/2)
which simplifies most of the things for you, infact i think you can do it with your hands tied behind your back.

-- AI
 
What is asked actually? to find r? Or something related to the triangle? If it is to find r, cos(2pi/3)=-1/2. the rest is easy.
 

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