- #1

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## Main Question or Discussion Point

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.

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

Thanks.

- Thread starter IndustriaL
- Start date

- #1

- 13

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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.

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

Thanks.

- #2

Integral

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Do you mean:

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

?

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

?

- #3

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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.

- #4

Gokul43201

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[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 ?

- #5

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Nope!Gokul43201 said:

[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 ?

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

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