Homework Help: Trigonometric and hyper. functions approx in large small argumnt

1. Mar 2, 2014

the_doors

hello guys ,

i'm looking for approximation of trigonometric and hyperbolic functions for small and large argument, is it correct to say sin(x)=x and tg(x)=x and tgh(x)=x and cos(x) = 1 and cosh(x)=1 and coth(x)=1/x for small x what about large x ? what can we say about exponential function in large and small argument ?

2. Mar 3, 2014

micromass

The key notion here is Taylor series. I don't think you've heard of that since you posted in precalc. But I think the answer to your question is going to be difficult without this notion.

The idea is that $\sin(x)\sim x$ for small values of $x$. Now, if $x$ becomes larger, then you need to add more and more error terms in order to still have a good approximation

For example, for somewhat larger $x$, we have

$$\sin(x)\sim x - \frac{x^3}{3!}$$

as a better approximation

See here for a list of approximations: http://en.wikipedia.org/wiki/Taylor_series#List_of_Maclaurin_series_of_some_common_functions
The more terms you use, the better the approximation. If you use infinite terms, then the approximation is usually exact.

3. Mar 3, 2014

HallsofIvy

Slightly simpler than Taylor's series here, but useful for these questions, is the "tangent line approximation". The derivative of sin(x) is cos(x) which is 1 at x= 0. Since sin(0)= 0 and its derivative at x= 0 is 1, the tangent line to the graph of y= sin(x) is y= 0+ (1)x= x. For small x, sin(x) is approximately equal to x.

Similarly tan(0)= 0, the derivative of tan(x) is sec^2(x), and sec(0)= 1 so the tangent approximation to y= tan(x) is also y= 0+ (1)x= x. (which, of course, means that for small x, tan(x) is approximately equal to sin(x).)

cos(0)= 1, the derivative of cos(x) is -sin(x), and -sin(0)= 0 so the tangent approximation to y= cos(x) is y= 1+ (0)x= 1.

Similarly, cosh(0)= 1, the derivative of cosh(x) is sinh(x), and sinh(0)= 0 so the tangent approximation to y= cosh(x) is y= 1+ (0)x= 1.

"1/x" is, of course, not a power series so not a Taylor's series. But it is true that tanh(0)= 1, the derivative of tanh(x) is sech^2(x), and sech(0)= 0 so that the tangent approximation to y= tanh(x) is y= x. Since coth(x)= 1/tanh(x), we can approximate y= coth(x) near x= 0 with 1/x.

Finally, for e^x, the derivative is e^x so both function and derivative are e^0= 1 at x= 0. For small x, y= e^x can be approximated by y= 1+ x.

e^x "goes to infinity" as x goes to infinity so we cannot say anything about an approximation to y= e^x for "large x".