Re: Range of Function

From: The World Wide Wade <waderameyxiii@xxxxxxxxxxxxxxxxxxxx>
Date: Thu, 29 Jun 2006 21:25:36 -0700

In article
<Pine.BSI.4.58.0606292045120.16292@xxxxxxxxxxxxxxxxx>,
William Elliot <marsh@xxxxxxxxxxxxxxxxxx> wrote:

On Thu, 29 Jun 2006, Simon Dean wrote:

I am trying to calculate the range of:

f(x) = sin (Arctan x)
-and-
g(x) = ln (cos x)

Any suggestions would be greatly appreciated!

sin arctan x = +-x/sqr(1 + x^2)

No, sin arctan x = x/sqr(1 + x^2). Besides, this step is a
needless complication.
.

References:
- Range of Function
  - From: Simon Dean
- Re: Range of Function
  - From: William Elliot

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