Comparing (sqrt(6)+sqrt(2))/2 and 5pi/6

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Maybe I'm just lazy, but it can also be solved in your head by approximating the values and using inequalities...
2<Sqrt(6)<3
1<sqrt(2)<2

From here:
5>sqrt(6)+sqrt(2)
Therefore:
(5/2=2.5)>(sqrt(6)+sqrt(2))/2

For the other side:
15<5pi
15/6<(5pi/6)

Therefore 5pi/6 > (sqrt(6)+sqrt(2))/2

rheymanda

cos(15) = cos(45 - 30) = cos(45)cos(-30) - sin(45)sin(-30) = cos(45)cos(30) + sin(45)sin(30)

And you can get an exact value for cos(15) from there, because 45-45-90 and 30-60-90 triangles are familiar.

morgancreighton

Easy enough. This is equivalent to determining the relation of 6sqrt(6)+6sqrt(2) to 10π. 2<sqrt(6)<3, so 12<6sqrt(6)<18. Similarly, 1<sqrt(2)<2, so 6<6sqrt(2)<12. Adding these inequalities yields 13<6sqrt(6)+6sqrt(2)<30. Meanwhile, 3<π<4, so 30<10π<40. We therefore have 6sqrt(6)<30<10π, meaning that (sqrt(6)+sqrt(2))/2<5π/6.

marcushendriksen

Just square both side. Left squared equals 2+sqrt(3)<4. Right squared >25*9/36>6. Obviously left<right.

ligion

This question should be something related to triangular function. 🤔🤔🤔

alextang

Nice problem!

I took a slightly different route, dividing both sides by 2 (placing everything in the first quadrant) and simplifying to get
1/√2 ( ½√3 + ½ ) ?? 5π/12
which can be re-expressed as
sin ¼π ⋅ cos ⅙π + cos ¼π ⋅ sin ⅙π ?? 5π/12
or
sin (¼π + ⅙π) = sin (5π/12) ?? 5π/12
to get a similar arc inequality result.

pietergeerkens

I solved it differently. I supposed (sqrt(6)+sqrt(2))/2<5pi/6. Cross multiplication and simplyfication give 3sqrt(6)+3sqrt(2)<5pi. Factorizing the first member gives 3sqrt(2)(sqrt(3)+1). This is surely less than 3sqrt(3)(sqrt(3)+1), which is 9+3sqrt(3). In turn 9+3sqrt(3) is less than 9+3sqrt(4), which is 9+6, or 15. SO far we have thus that 3sqrt(6)+3sqrt(2)<15. 5pi is now > 5x3=15. So, closing the chain, we have 3sqrt(6)+3sqrt(2)<15<5pi. So the initial supposition was correct and (sqrt(6)+sqrt(2))/2<5pi/6

andreabaldacci

The geometric demonstration is quite unnecessary here. For any x > 0 we have

sin(x) < x

Consequently, we have

sin(⁵⁄₁₂π) < ⁵⁄₁₂π

and since

sin(⁵⁄₁₂π) = sin(¼π + ⅙π) = sin(¼π)·cos(⅙π) + cos(¼π)·sin(⅙π) = ½√2·½√3 + ½√2·½ = (√6 + √2)/4

it follows that

(√6 + √2)/4 < ⁵⁄₁₂π

and therefore

(√6 + √2)/2 < ⅚π

NadiehFan

Have you ever done one of these problems where they’re actually equal? It’d be a twist ending!!!

stvp

We can solve it by comparing the both numbers with 2.5 and will get the result

sourajitdas

If you could come up with a problem that depends on the fact that π < √10, that might be interesting. π² = 9, 87, so it's pretty close. Maybe something related to π/10 cords, sin, etc., would work out. I tried a couple of things and couldn't get there.

mbmillermo

Well. 2.5^2=6.25
That mean sqrt(6) and sqrt(2) is less than 2.5... that mean the left side of the eqation is less than (<2.5+<2.5)/2=<2.5 . We know pi is larger than 3 so the right side of the eqation is more than 2.5.. (3*5)/6=2.5 therefore the right side is larger

domahidipeter