Is there an actual formula for finding Sin, Cos, Tan, etc.?
If not, then how do calculators find it?
I don't understand what you're saying.
Calculators take a taylor series approximation that's good enough for anything.
>>7650963
[math]tan(x)=sin(x)/cos(x)[/math]
[math]cos(x)=sin(x)*tan(x)[/math]
[math]sin(x)=cos(x)*tan(x)[/math]
:^)
[math]\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}[/math]
[math]\cos(x) = \frac{e^{ix} + e^{-ix}}{2}[/math]
[math]\tan(x) = \frac{\sin(x)}{\cos(x)}[/math]
Calculators use lookup tables with approximative calculated values.
SOH
CAH
TOA
>>7650976
great job not understanding the question at all you fuck
Look up taylor series. Basically, they can approximate the sine wave as a complicated polynomial equation. The more accurate you want the answer to be, the more terms you can add to the polynomial
>>7650977
Calm down son
tanx = sinx/cosx
cosx = sqrt(1-(sin^2)x)
calculate sinx using the equation six = (sinx)/n
>>7650963
cordic algorithm
>>7650963
[math]\sum_{i=0}^{\infty} \frac{(-1)^i \theta^{2 i+1}}{(2 i+1)!} [/math]
test
>>7650963
look you 14 year old,
sin x = x^2 - 1/3 x^4 etc. Look up power series for sine.
>>7650967
Taylor series are horrible inefficient in certain type of hardware. Calculators often use the CORDIC algorithm.
e^x = sum n = 0; n -> inf (1/n!) * x^n
e^jx = cos x * j sin x
(j is complex)
Figure the rest for yourself.
>>7652236
>[/eqn]
maybe this
[math]4^{(1/x)-1}\cos\frac\pi x\
=\frac14+\frac{x-4}{2!x^2}-\frac{(x-4)(2x-4)(3x-4)}{4!x^4}+\frac{(x-4)(2x-4)(3x-4)(4x-4)(5x-4)}{6!x^6}\mp\ldots[/math]
>>7652238
I've tried like three times just gonna post a pic, fuck tex
>>7651438
>>7651295
>>7650967
>>7650978
>2015
>using Taylor series
>Not making your own series that approximates trigonometric functions without any transcendental numbers
The beauty of my formula is that you can for some cases actually get closed-form solutions out of it instead of an endless string of pis. Try x= 4 or x= 2 etc.
does this shit still work?
[math]2^{2}[/math]
>>7652247
apparently not
>>7652588
TEX is broken. Something to do with the whitespace, for whatever reason. We need to complain to an Hiro.
The problem is pretty weird. One anon tried figuring it out a few weeks back, and I tried to help a bit, but he couldn't do it.
>>7652603
Sorry, 4chan's LaTeX rendering is broken. But the preview works fine.
[math]\sum_{i=0}^{\infty} \frac{(-1)^i \theta^{2 i+1}}{(2 i+1)!}[/math]
my father used to have a booklet with values of sin, cos, maybe tan too, to use at school. Since they did not have calculators.
Someone give me a proper rundown of this
>>7652981
offical maths definition is taylor series
calculators use algorithms however because its more efficient
If you want closed form solutions the only equation that can do it is the series i invented but nobody cares about that.
sin and cos are defined by e, which is a transcendental number. So you can't "find" sin or cos in the same way you can't find the last digit of pi
>>7653071
this is completely untrue. The exponential formula for trigonometric ratios are evaluated in the complex plane where transcendental functions can reduce to algebraic ones. Does e^i*pi equal a transcendental number
Huh, this place has TeX?
[eqn]
\frac{huh}{kay}
[/eqn]
>>7653048
I'm a math noob. Explain how to use your equation senpai.
>>7654668
uh well if you want to evaluate cos (pi/7) make x = 7 for the series.
>>7655700
And how could i apply this to practical situations?
>>7653109
e is transcendental, therefore [math] e^{i \theta } [/math] is transcendental unless [math] \theta [/math] is also transcendental. And yes I'm implying 0 is transcendental.
>>7650963
I can't offer any mathematics, but Wikipedia will lead you to the historical derivations that mathematiclans invenvented when trigonometry was a new field. This of course will be horribly inaccurate in today's digital world yet it will give you a solid understanding of why trigonometry operates the way it does.
>>7655700
How is it closed form if you use an infinite series. What do you mean by closed form solution?
>>7650963
[math]\sin(\pi z) = \pi z \prod_{n=1}^{\infty} \left(1 - \frac{z^2}{n^2}\right) = \frac{\pi z}{ z! * (-z)! } [/math]
Where z! = z*(z-1)*...3*2*1
>>7656606
>defining factorial but not that weird ass giant pi summation symbol
thanks senpai
>>7656607
>defining factorial but not that weird ass giant pi summation symbol
Do people really not learn about products in grade school?
∏f(n)=exp(∑Log(f(n)))
Happy?
[math]\cos(x)= \frac{2}{\pi} \sum_{n=2}^{\infty} \frac{ n( (-1)^n+1)} {n^2-1} \sin( n x )[/math]
>>7656640
Not quite
>>7650963
lookup table
>>7656638
Excuse me? Infinite series is not something you learn in high school. Nice try at fitting in with all the other cool kids on /sci/ by pushing the high school meme though!
>>7651452
This desu senpai
>>7656853
infinite series is precalc dude.
>>7657340
infinite series is calc 2 you retard
>>7652247
>[math]i^{i}=0.21 \dots[/math]
Source?
>>7650963
Chebyshev polynomials
and in the case of say IEEE 754 floating point below a certain threshold sin(x) = x (common optimization)
>>7657348
At shitty schools perhaps. Good schools do it in precalc
>>7657351
i^i= (e^iπ/2+2πn*i)^i=(e^-π/2+2πn)= 0.0216069591 (for n=0)
>>7657353
find a school that does infinite series in precalc.
>>7657351
[math] e^{i \pi } =-1 [/math]
[math] \sqrt{e^{i \pi }} = \sqrt{-1} [/math]
[math] e^{i \frac{ \pi}{2} } =i [/math]
[math] e^{i \frac{ \pi}{2} }^i =i^i [/math]
[math] e^{i^2 \frac{ \pi}{2} } =i^i [/math]
[math] e^{- \frac{ \pi}{2} } =i^i [/math]
>>7657367
That's the way the subject was original taught. Euler does it in his precalc book
>>7656585
What is convergence?
>>7657373
Interesting, thank you!
As someone else has said, CORDIC is one way to do it and is a clever low needs method.
Taylor series are actually NOT the way most calculators will find solutions. They should be programmed with a minimax approximation which is similar to a Taylor series polynomial, but the coefficients are such that the maximum error is minimized throughout the entire function. This is better than a Taylor series because Taylor series have increased error as the function goes away from the central point.
You also of course get to use tricks from the periodicity of the functions, you really just need to have the first quarter of the curve done well because it can be flipped around to make everything else.
>>7650963
http://intmstat.com/blog/2011/06/exact-values-sin-degrees.pdf
>>7657367
Can confirm, was taught this in precalc.
>>7657385
>programmed with a minimax approximation
the search algorithm?
>>7656585
ayy lmao guys if you put x=4 into my series every term past the first one will equal zero therefore you are left with a closed form solution. Get it now?
>>7657377
No I am saying my series doesnt always have to be summed to convergence, it can simply reduce to a closed expression because all the terms past the first one disappear in some cases. See above.
>>7657438
No, there are a confusingly large number of things called "minimax". The Stone–Weierstrass theorem guarantees a solution exists, and you would use something like Remez's algorithm to find the polynomial. Another alternative is to use Chebyshev polynomials, I believe they fall into the "good enough" category. Info on approximations: http://dlmf.nist.gov/3.11
>>7657367
Infinite series are covered yes, but butchered. You only learn their anatomy and little properties, nothing like what you do in calc.
>>7650963
http://amycoders.org/tutorials/sintables.html
>>7657373
Why is it implied that i^2 is -1 ?
>>7659307
Because [math]\sqrt{-1} = i[/math]
>>7657340
Okay, so proove that infinite seria of continuous functions is continuous too
>>7659318
You're wrong.
[math]\sqrt{-1} = \pm i[/math]
>>7651014
What is cordic?
I prefer taylor series
>>7650971
>lookup table
Wouldn't that be costly, considering the precision?
>>7657353
>HURR I CONSTRUCTED TAYLOR SERIES APPROXIMATIONS WITH SPECTRAL ACCURACY IN KINDERGARTEN. LE GIT GUD MEME.
>>7653048
>If you want closed form solutions the only equation that can do it is the series i invented
idontbelieveyou.gif
>>7656853
this has to depend on country desu senpai, we did this in high school for sure
>>7659305
Not either of them, but they were calc 2 for me.
>>7659372
>so proove that infinite seria of continuous functions is continuous too
For that last time Cauchy it isn't. See the Fourier series of a square wave.
>>7652619
[math]It takes literally 5 minutes to learn.[/math]
>>7660593
But it some sense it's correct. Consider a sequence [math](f_k)[/math] of functions with [math]f_k \in C(\mathbb{R})[/math] for all [math]k \in \mathbb{N}[/math].
Suppose that the series [math]\sum_{k=0}^\infty f_k = \lim_{N \to \infty} \sum_{k=0}^N f_k[/math] converges to [math]f[/math] then [math]f \in C(\mathbb{R})[/math].
But if you only have that [math]\sum_{k=0}^\infty f_k(x) = \lim_{N \to \infty} \sum_{k=0}^N f_k(x) = f(x)[/math] for all [math]x \in \mathbb{R}[/math] then it's not necessary that [math]f \in C(\mathbb{R})[/math],
The difference is that for convergence you always use the norm of the underlying space which is the supremum norm in [math]C(\mathbb{R})[/math] (uniform convergence) while convergence for all inputs is about convergence in [math]\mathbb{R}[/math] (pointwise convergence).
>>7660601
[math]\frac{\frac{\frac{\frac{i}{just}}{use}}{a}}{chart}[/math]
>>7656619
kek, nice definition
>>7660619
Why is this so inconsistent?
>>7660619
>[math] \frac{ \frac{ \frac{ \frac {I} {just} } {use} } {a} } {chart} [/math]
>>7657348
[math] i^i = {\left( e^{i\frac{\pi}{2}} \right)}^i = e^{-\frac{\pi}{2}} \approx 0.21 [\math]
>>7660692
[math] i^i = {\left( e^{i\frac{\pi}{2}} \right)}^i = e^{-\frac{\pi}{2}} \approx 0.21 [/math]
>>7660686
Did you literally just mark and quote that?
And that worked?
>>7650971
How you gonna find [math]e^{ix}[\math] if you don't have sin and cos already
>>7660704
No, he liturgically marked and quoted it.
>>7660715
I don't think you're using that word correctly.
Unless perhaps he prayed to 4chan to let it work.
>>7660707
[eqn]e^{ix} = \sum_{k=0}^\infty \frac{(ix)^k}{k!} [/eqn]
>>7660729
So.... your plan to compute tan(x) is to just compute four seperate taylor series expansions?
When there already exists a single taylor series expansion you could use for tan(x)?
Not to mention, better algorithms than taylor series expansions?
>>7660732
OP's question was whether an formula exists not whether it's practical.
>>7660601
No, brah, there is something wrong with it. It glitches out sometimes and doesn't render. I wish they would just hand it off to regular ol' mathjax or katex or something instead of the stupid shit they have [math]trying[/math] to render it
>>7660619
I see you have an egg on your face there
>>7660588
Nice meme but you will never hold a candle to me in number theory.
>>7660723
He's one of those faggots who gets an aneurysm whenever they see someone using "literally"
>[math]\frac{this is}{a test}
[/math]
>>7661021
Okay, so it doesn't work like /jp/'s Japanese text thing.
>[math]\frac{this is}{a test}[/math]
>>7661022
> [math]\frac{stealing}{yourtest}[/math]
>>7661037
Someone tell me what i'm doing wrong.
>>7661037
>[math]\frac{it's~probably}{the~space}[/math]
>>7661058
>[math]\frac{it is probably}{the space}[/math]
>>7661061
>[math]\frac{I~it~was}{the~apostrophe}[/math]
>>7661062
>[math]\frac{boobieboobie}{bumbum}[/math]
[math]\frac{test}{other}[/math]
>>7659573
In 1970s maybe.
>>7662278
Yo what the actual shit? I posted this on my stationary computer and it didn't show as working. Now i'm on a laptop elsewhere and it works just fine.
