/sci/ - Science & Math » Thread #11862115

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Anonymous Fri 03 Jul 01:27:23 2020 No.11862115 View Reply Original Report

Quoted By: >>11862125 >>11863631 >>11863939 >>11864017

u should be able 2 solve this

Anonymous

Anonymous Fri 03 Jul 2020 01:29:37 No.11862120 Report

Quoted By: >>11862122

3 sqrt 2 + 1? So approx 5.24

Anonymous

Anonymous Fri 03 Jul 2020 01:30:40 No.11862122 Report

Quoted By:

>>11862120
Wait 3 sqrt 2 + 2, forgot about the cot

Anonymous

Anonymous Fri 03 Jul 2020 01:31:09 No.11862124 Report

Quoted By: >>11862128

what is t?

Anonymous

Anonymous Fri 03 Jul 2020 01:31:13 No.11862125 Report

Quoted By: >>11863523

>>11862115
Engineer here.
sin t = t
cos t = 1
tan t = t
cot t = 1/t
sec t = 1
csc t = 1/t

Hence the answer is 2

Anonymous

Anonymous Fri 03 Jul 2020 01:32:15 No.11862128 Report

Quoted By:

>>11862124
a Real number

Anonymous

Anonymous Fri 03 Jul 2020 11:23:11 No.11863047 Report

Quoted By: >>11863055

Approximately 1.828427, right?

Anonymous

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Anonymous Fri 03 Jul 2020 11:37:17 No.11863055 Report

Quoted By:

>>11863047
That's also what I got.

Anonymous

Anonymous Fri 03 Jul 2020 16:19:21 No.11863523 Report

Quoted By:

>>11862125
based approximation

Anonymous

Anonymous Fri 03 Jul 2020 16:54:34 No.11863625 Report

Quoted By:

two times the square root of two, minus one.

Anonymous

Anonymous Fri 03 Jul 2020 16:57:58 No.11863631 Report

Quoted By:

>>11862115
infinity/t * abs(...) = infinity/t = infinity*sign(t)

Anonymous

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Anonymous Fri 03 Jul 2020 18:42:31 No.11863939 Report

Quoted By:

>>11862115
$\inf \left(\cot t + \sec t = \frac{\cos t + 1}{\sin t} \right) = - \infty$ around 0. This is also the final result, since all other terms are bounded around 0.

Anonymous

Anonymous Fri 03 Jul 2020 19:11:47 No.11864017 Report

Quoted By:

>>11862115
There is also a trick to solve these kinds of problems.
Let
$ x = \tan\left( \frac{t}{2} \right) $
then
$ \sin(t) = \frac{2x}{1+x^2} \cos(t) = \frac{1-x^2}{1+x^2} \tan(t) = \frac{2x}{1-x^2} \cot(t) = \frac{1-x^2}{2x} \sec(t) = \frac{1+x^2}{1-x^2} \csc(t) = \frac{1+x^2}{2x} $

So we just need to find the infinmum of
$ \inf_x \left| \frac{2x}{1+x^2} + \frac{1-x^2}{1+x^2} + \frac{2x}{1-x^2} + \frac{1-x^2}{2x} + \frac{1+x^2}{2x} \right| = \inf_x \left| -2 + \frac{1}{x} - \frac{2}{x-1} + \frac{2x+2}{x^2+1} \right| $
Now let
$f(x) = 2 + \frac{1}{x} - \frac{2}{x-1} + \frac{2x+2}{x^2+1}$
then
$f'(x) = 0$
has the solutions
$ x_1 = \sqrt{2} - 1 \\ x_2 = -1 - \sqrt{2} \\ x_3 = 1 + \frac{1}{2} \sqrt{2} + \frac{1}{2} \sqrt{10 + 8 \sqrt{2}} \\ x_4 = 1 + \frac{1}{2} \sqrt{2} - \frac{1}{2} \sqrt{10 + 8 \sqrt{2}} \\ x_5 = 1 - \frac{1}{2} \sqrt{2} + \frac{1}{2} \sqrt{10 + 8 \sqrt{2}} \\ x_6 = 1 - \frac{1}{2} \sqrt{2} - \frac{1}{2} \sqrt{10 + 8 \sqrt{2}} $
For each of those solution you can check
$|f(x_i)|$
to find that the infinimum is
$|f(x_3)| = 2 \sqrt{2} - 1 \approx 1.828427$

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