Can Veeky Forums do elementary high school trig: the thread

how do retards like this even breathe

x = ± 1

bump

x = +- sqrt( pi^2 n^2 + 1) - pi n
where n is an integer

x = -z*π ± sqrt( (z*π)^2 + 1)

where z is any integer

ah, beat me to it

We wish to solve [math]\sin \frac1x = \sin x[/math]. Because [math]\sin(x + 2\pi) = \sin x = \sin \bigl( \frac\pi2 - x \bigr)[/math], there are two cases: [math]\frac1x = x + 2 \pi k[/math] and [math]\frac1x = \frac\pi2 - x + 2 \pi k[/math], both of which are for arbitrary [math]k \in \mathbb Z[/math].

In the first case, WLOG [math]x^2 - 2\pi k x - 1 = 0[/math], which we solve to find [math]x = \pi k \pm \sqrt{\pi^2k^2 - 1}[/math]. The second case is similar and more disgusting. So fuck writing that out.

Also, Wolfram|Alpha would have done this homework for you just as well: wolframalpha.com/input/?i=sin(x) = sin(1/x)

>difficult
please go away

x = -sqrt(π^2 n^2+1)-π n, n element Z

I don't remember doing this in high school

Sin(1/x) = sin(x)
Sin^-1(sin(1/x)) = sin^-1(sin(x))
1/x =x
1 = x^2
+/-1 = x

>[math]\sin[/math] is invertible
>look see i inverted it: [math]\sin^{-1}[/math]
you might wanna retake precalc there buddy

WOW you dont even know inverse from invert!

I love it, keep going

Sin(x)=y
Arcsin(y)=x

Does your calculator say arcsin or sin^-1 ??

Okay

>Inverse trigonometric functions
>From Wikipedia, the free encyclopedia
> (Redirected from Arcsine)
>
>In mathematics, the inverse trigonometric functions (occasionally called cyclometric functions[1]) are the inverse functions of the trigonometric functions (with suitably restricted domains).

>with suitably restricted domains

Is that enough or should I go on?

Im glad you finally learned something

wow teach thanks, i just have one more question
how do i not lose solutions to the original equation when i restrict the domain of [math]x[/math]?
because this is a lossy solution method, of course

Add n(pi)

2n*pi you fucking pleb

I'm more interested in all possible answers.

sin(1/x) will approach 0 as x goes to infinity, and sin(x) will intersect these values for an infinite amount of values, although they will be slightly further apart each time. The separation of values will approach the finite limit of pi, but what values do they take before the limit?

sin(x)=sin(pi-x) not pi/2 so
1/x=2kpi+x or (1+2k)pi-x
Also your solution should hold for k=0, it should be +1 in the root.

2pi gets you to the same place!

I really like seeing you try, please continue!