Which is larger? The difference is less than 0.001

This is a tricky no-calculator question. Which is larger: pi/12 or a nested radical equal to sqrt(2 - sqrt(2 + sqrt(3))) ?

@SyberMath

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[PoppySuzumi]

A good advanced example to learn Trigonometry.

[maxhagenauer]

You can do it knowing the method of computing pi with an infinite nested square roots from polygons approaching circles.

[randomjin]

We have √(2+√3) = 2√((1+√3/2)/2) = 2√((1+cos(𝜋/6))/2) = 2cos(𝜋/12) Next, √(2-√(2+√3)) = √(2-2cos(𝜋/12)) = 2√((1-cos(𝜋/12))/2) = 2sin(𝜋/24). Knowing that for all x > 0, sin(x) < x, we have sin(𝜋/24) < 𝜋/24, so 2sin(𝜋/24) < 𝜋/12 meaning √(2-√(2+√3)) < 𝜋/12

[ZijZijnZijnZoons]

I appreciate the great content. Sometimes I cannot easily follow it, but it makes me think out problems more.

[ExtraMedium-]

If the difference is less than 0.001, then they are equal.
Sincerely,
A scientist

[bjorneriksson]

Well, that's lame, realizing that it represented the relationship between a chord and an arc! As any real man does, I chose to completely overlook that clever and insightful solution and instead brute force calculate the whole square root expression in my head, with up to 5 decimal places. I (eveeentually) came up with the answer that the square root expression was almost exactly 0.261 and that pi/12 was about 0.2618. The calculator then confirmed this was indeed correct (0.26105 < 0.2618)! I was right, what a delight! 😄

[deerho]

I think the key is analyzing that what makes you compare an expression with pi with an expression that has square roots, especially root 3. That leads naturally to trig, and clearly to start with an angle of pi/12 which leads to the 1/2 angle formula. But it is a contrived problem based on the trig and circle relationship and not something that would occur organically or as an outshoot of another exploration.

[marcusscience]

Constructing a perpendicular bisector of line c, passing through the center of the arc and also bisecting angle θ, we get the formula:
c = 2 * sin(θ/2)
For smaller and smaller angles of θ, the difference between θ and c will approach 0, and looking at the diagram it’s easy to see why.

[AmlanSarkar-wrpr]

Legends assuming π = 180° and saying π/12 is large 💀💀

[RyanK-]

TEST TAKING HINT: The larger quantity is almost always the one with the pi. The question writers like these kinds of problems. The arc length will have a pi and the smaller segment will not. Just choose the pi and move on with the test. Come back to this one if there's time.

[henrymarkson]

I once wrote Theorem of Al-Kashi in a test.
Teacher crossed it out and wrote Law of Cosines.
I failed the test because of that.
I'll never make that mistake again, it was a valuable lesson.

[jacobgoldman]

Shouldn’t we more thoroughly prove that for all theta c<theta?

[MCLastUsername]

This seems like more of a mind-reading problem than a math problem

[UmarFarooq-bbqs]

Here is how I did it:
Take a unit circle. Draw a sector with an angle of 30deg. Inscribe a triangle in that sector. The arc length of that sector is π/12. Calculate the length of the base (the side whose endpoints intersect the circumference) of that triangle using the cosine rule. It comes out to be sqrt(2 - sqrt (3)). Compare this expression with the one on the RHS in the question: on simple comparison, it can be observed that it is obviously bigger. However, since the arc of a circle is longer than a straight line for the same subtended angle, we can conclude that the LHS is bigger than the RHS.

[augtsu]

Between this example then apply it to the interesting fact that the two towers on the Golden Gate Bridge are out of parallel by about 8" (if memory serves me correctly). This gives me the answer when a child asks that proverbial "Why do I need to learn this?" Thanks for sharing.

[High-Tech-Geek]

To solve this you have to be intimately aware of standard theta value representations and cosines of pi.
I am not, so I didn't recognize the initial relationship between the 2 samples given.
I feel like this problem was kind of created through reverse engineering of sorts, haha.

[Pseudoaesthetic]

Calculator enjoying the content in a corner with puffs 😂😂😂

[JonathanKahan-qg]

I had to use the brute force approach where i used the Pythagorean theorem in repeated application. First to compute the chord when it is 30 degrees, and then again for 15 degrees. Took me much longer than you guys but did not need any trig.

[MustafaKhan-hzmr]

i havent watched the video but does it involve splitting a circle into 12 parts and drawing a triangle in one of the parts to approximate the area

[itsphoenixingtime]

I considered one sector of a 24-gon and calculated from there. I had struggles with the square root though, as I kept using the different of angles law and not the double angle formula. So that was my mistake from there as I should have double angled pi/6 twice.