sin cos tan

@PangBrownBear said in #8:
> sinh?

sinh and cosh are like the Sys Rq key on your keyboard. No one knows what they're for and no one ever uses them, but they're there. tanh is like the Insert key. Most people never use them, but I do on occasion (for overwriting text in the case of Insert, and for machine learning in the case of tanh).

@AsDaGo said in #11:
> sinh and cosh are like the Sys Rq key on your keyboard. No one knows what they're for and no one ever uses them, but they're there. tanh is like the Insert key. Most people never use them, but I do on occasion (for overwriting text in the case of Insert, and for machine learning in the case of tanh).
oh, got it, ty

sohcahtoa

also, is y = cosh(x) a true parabola?

It's quite common in India to learn hard chapters which are almost 2 to 3 grades higher.Kind of trend here for the smart folks.

@MIHIR_KATTI said in #14:
> also, is y = cosh(x) a true parabola?

No, while some portions of cosh(x) near x = 0 might closely resemble a parabola (this is not a coincidence due to the presence of a term of order 2 in the Taylor expansion of cosh(x)), the further you get from x = 0 the more important the contribution of higher order terms becomes. That is, the graph gets way too steep way too quickly for a parabola due to the (x^4)/24 term in its expansion.

If it were a parabola its (finite) Taylor series would be of order 2 (and all higher order terms would vanish). Instead its Taylor series is an infinite series (containing terms of arbitrarily high order):
www.wolframalpha.com/input?i=taylor+series+cosh%28x%29+at+x%3D0

So it cannot be a parabola. You can also see this directly from the definition of cosh:
cosh(x) = (e^x + e^(-x))/2

For large x-values cosh(x) approaches (1/2)*e^x asymptotically. That is, it behaves more and more like the exponential function (which is not a polynomial, so it cannot be a polynomial of order 2).

@Thalassokrator said in #16:
> No, while some portions of cosh(x) near x = 0 might closely resemble a parabola (this is not a coincidence due to the presence of a term of order 2 in the Taylor expansion of cosh(x)), the further you get from x = 0 the more important the contribution of higher order terms becomes. That is, the graph gets way too steep way too quickly for a parabola due to the (x^4)/24 term in its expansion.
>
> If it were a parabola its (finite) Taylor series would be of order 2 (and all higher order terms would vanish). Instead its Taylor series is an infinite series (containing terms of arbitrarily high order):
> www.wolframalpha.com/input?i=taylor+series+cosh%28x%29+at+x%3D0
>
> So it cannot be a parabola. You can also see this directly from the definition of cosh:
> cosh(x) = (e^x + e^(-x))/2
>
> For large x-values cosh(x) approaches (1/2)*e^x asymptotically. That is, it behaves more and more like the exponential function (which is not a polynomial, so it cannot be a polynomial of order 2).
wow, what an essay!

@PangBrownBear said in #1:
> Anyone learned trigonometry who is under 13?

First of all, no one under 13 should be allowed on this web site. There is too much adult content here that would confuse you.
And as for trigonometry, it is more important these days for people to learn about fashion than trigonometry. I mean just look around at those who play chess here. Except for me, no one seems to know the first thing about fashion. Don't you agree?

@Thalassokrator said in #16:
> No, while some portions of cosh(x) near x = 0 might closely resemble a parabola (this is not a coincidence due to the presence of a term of order 2 in the Taylor expansion of cosh(x)), the further you get from x = 0 the more important the contribution of higher order terms becomes. That is, the graph gets way too steep way too quickly for a parabola due to the (x^4)/24 term in its expansion.
>
> If it were a parabola its (finite) Taylor series would be of order 2 (and all higher order terms would vanish). Instead its Taylor series is an infinite series (containing terms of arbitrarily high order):
> www.wolframalpha.com/input?i=taylor+series+cosh%28x%29+at+x%3D0
>
> So it cannot be a parabola. You can also see this directly from the definition of cosh:
> cosh(x) = (e^x + e^(-x))/2
>
> For large x-values cosh(x) approaches (1/2)*e^x asymptotically. That is, it behaves more and more like the exponential function (which is not a polynomial, so it cannot be a polynomial of order 2).
but we could do y = cosh(x) + m whereas m is the y intercept. The variable m proves that the function does not have to intercept at 0.

@PangBrownBear said in #1:
> Anyone learned trigonometry who is under 13?

I was like 12 at that time