/sci/ - Science & Math » Thread #13593011

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Anonymous Wed 01 Sep 23:42:21 2021 No.13593011 View Reply Original Report

Quoted By: >>13595947 >>13596013 >>13596371

$\sin^2x+\cos^2x=1, \forall x\in \mathbb{R}$
Never forget this trig identity, it carried us all in basic trig and still in advanced trig.

Anonymous

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Anonymous Wed 01 Sep 2021 23:50:02 No.13593041 Report

Quoted By: >>13593126

>*creates the ability measure distances*
enjoy this invention of mine! you'll need it.

Anonymous

Anonymous Thu 02 Sep 2021 00:15:45 No.13593126 Report

Quoted By: >>13596153

>>13593041
Thanks for taking the naming credit of something invented by the Babylonians

Anonymous

Anonymous Thu 02 Sep 2021 20:02:21 No.13595947 Report

Quoted By:

>>13593011
why do I need this?
just give me $\sin{x}$ apples lmao

Anonymous

Anonymous Thu 02 Sep 2021 20:20:35 No.13596013 Report

Quoted By: >>13596048 >>13596371 >>13596643

>>13593011
I also like
$\cot^2x+1=\csc^2x, \forall x \in \mathbb{R}$

Anonymous

Anonymous Thu 02 Sep 2021 20:33:06 No.13596048 Report

Quoted By: >>13596055 >>13596412

>>13596013
Just divide through by $\sin^2 x$

Anonymous

Anonymous Thu 02 Sep 2021 20:35:54 No.13596055 Report

Quoted By: >>13596060

>>13596048
I'll do it again
$\frac{cot^2x}{sin^2x}+\csc^2x=\csc^4x$

Anonymous

Anonymous Thu 02 Sep 2021 20:37:21 No.13596060 Report

Quoted By: >>13596070

>>13596055
I meant the original equation, but this is a truly astounding gem of trigonometry

Anonymous

Anonymous Thu 02 Sep 2021 20:41:25 No.13596070 Report

Quoted By: >>13596161

>>13596060
I shall keep going
$\frac{\cot^2x}{\sin^4x}+\csc^4x=\csc^6x$

Anonymous

Anonymous Thu 02 Sep 2021 21:09:33 No.13596153 Report

Quoted By:

>>13593126
Indeed. And equally, since there once was a Babylonian who discovered
that when an apple detaches from a tree, it usually falls to the ground, it
means that Newton stole the entire theory of classical mechanics from them
as well.

Anonymous

Anonymous Thu 02 Sep 2021 21:13:32 No.13596161 Report

Quoted By: >>13596173

>>13596070
Can anything stop him ?

Anonymous

Anonymous Thu 02 Sep 2021 21:19:38 No.13596173 Report

Quoted By: >>13601754

>>13596161
$\frac{\cot^2x}{\sin^6x}+\csc^6x=\csc^8x$

Anonymous

Anonymous Thu 02 Sep 2021 22:47:48 No.13596371 Report

Quoted By: >>13596412 >>13596418

>>13593011
this describes points on a unit circle

[sin(x), cos(x)]

which have a magnitude defined by your equation

So what would
>>13596013
mean?

btw how do i into latex on 4chanz

Anonymous

Anonymous Thu 02 Sep 2021 23:04:35 No.13596412 Report

Quoted By: >>13596524

>>13596371
$\cot^2x+1=\csc^2x$
Is just an identity, you can achieve this equation by dividing through by $\sin^2x$ as >>13596048 says

Anonymous

Anonymous Thu 02 Sep 2021 23:08:22 No.13596418 Report

Quoted By:

>>13596371
You can use latex by using " $" and ["/math"]$

Anonymous

Anonymous Thu 02 Sep 2021 23:55:00 No.13596524 Report

Quoted By: >>13596608

>>13596412
ok so
$cot^2 x + 1 = csc^2 x \implies |\begin{bmatrix}cot x \\<br />i csc x\end{bmatrix}|=-1$

And I'm thinking $i cos(x) = sin(x) \implies i cscx = -secx$ .

so the graph would be something like this

https://www.math3d.org/gbBijuPC

Anonymous

Anonymous Fri 03 Sep 2021 00:19:41 No.13596608 Report

Quoted By: >>13596624

>>13596524
Congratulations, you plotted the unit circle and three of the four unit hyperbolae.

Anonymous

Anonymous Fri 03 Sep 2021 00:22:52 No.13596624 Report

Quoted By:

>>13596608
thanks
give me more identities and i will keep going

Anonymous

Anonymous Fri 03 Sep 2021 00:28:21 No.13596643 Report

Quoted By: >>13596979

>>13596013
> $\forall x\in \mathbb{R}$
I'm sure this will work for $x=0$

Anonymous

Anonymous Fri 03 Sep 2021 01:52:55 No.13596979 Report

Quoted By: >>13598169

>>13596643
I am so dumb, $\forall x \in \mathbb{R}, x \neq 0$

Anonymous

Anonymous Fri 03 Sep 2021 09:54:45 No.13598169 Report

Quoted By: >>13598678

>>13596979
I'm sure it will also work for $x=\pi$

Anonymous

Anonymous Fri 03 Sep 2021 12:48:58 No.13598575 Report

Quoted By: >>13599557

Memorizing the complex exponentials is also incredibly useful, even though for some reason it is not taught in calculus.

Anonymous

Anonymous Fri 03 Sep 2021 13:16:20 No.13598678 Report

Quoted By:

>>13598169
Having only a countable number of exceptions is still insignificant.

Anonymous

Anonymous Fri 03 Sep 2021 17:05:31 No.13599557 Report

Quoted By: >>13599927

>>13598575
Complex exponentials, plural?
You mean like $\cosh(iz) = cos(z)$

Anonymous

Anonymous Fri 03 Sep 2021 18:23:06 No.13599927 Report

Quoted By: >>13599941

>>13599557
No, these
$<br />\cos x = \frac{1}{2}(e^{ix} + e^{-ix}), \enspace \sin x = \frac{1}{2i}(e^{ix} - e^{-ix})<br />$
etc.. makes it trivial to modify or derive a bunch of trig ids when you need them.

Anonymous

Anonymous Fri 03 Sep 2021 18:25:31 No.13599941 Report

Quoted By:

>>13599927
Or even cooler
$e^{ix} = cos(x) + isin(x)$

Anonymous

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Anonymous Fri 03 Sep 2021 19:35:51 No.13600158 Report

Quoted By:

$J^3=j^2J\ \implies\ \exp(z\,J) = E + \dfrac{\sinh(z\,j)}{j} J + \dfrac{\cosh(z\,j)-1}{j^2} J^2$

Anonymous

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Anonymous Fri 03 Sep 2021 19:41:57 No.13600180 Report

Quoted By:

For 3x3 anty-symmetric matrices, that's

https://en.wikipedia.org/wiki/Rodrigues%27_rotation_formula#Matrix_notation

pic related. More broadly

$\sum_{k=0}^\infty a_k J^k = a_0\,E + \sum_{i=0}^\infty \left(a_{2i+1}\, E + a_{2i+2}\, J\right)\,(j^{2})^iJ$

but if you don't go with z^k/k! then the connection to the trig identity is entirely lost as far as recognizability goes.

Anonymous

Anonymous Sat 04 Sep 2021 08:24:43 No.13601754 Report

Quoted By:

>>13596173
kek

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