NetNews Usenet Archive 1992 #27/NN_1992_27.iso/spool/rec/puzzles/7446

Internet Message Format | 1992-11-22 | 2.0 KB

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From: [e-mail] (Carl J Lydick)
Newsgroups: rec.puzzles
Subject: Re: Trig. inequality
Date: 22 Nov 1992 16:26:40 GMT
Organization: HST Wide Field/Planetary Camera
Lines: 24
Distribution: world
Message-ID: <[e-mail]>
References: <[e-mail]>,<[e-mail]>
Reply-To: [e-mail]
NNTP-Posting-Host: sol1.gps.caltech.edu

In article <[e-mail]>, [e-mail] (J. Theodore Schuerzinger) writes:
=> Show that (sin x)^(sin x) < (cos x)^(cos x) when 0 < x < pi/4.
=
=Simple math.  Remember that (sin x)^2+(cos x)^2=1.  An identity from my
=trig class back in high school.  Setting sin x equal to cos x, we get
=(sin x)^2=1/2.  The answer to this is x=pi/4.  And since sin x
=increases from 0 to 1 as x increases from 0 to pi/2, it's obvious that
=cos x will be greater than sin x.  As neither is negative in this
=range, (cos x)^2 will be greater than (sin x)^2 from 0<x<pi/4.  I
=haven't taken any math since freshman year of college, but this was
=easy!

And wrong.  Fine; you've proven that sin(x)^2 < cos(x)^2 for 0<x<pi/4. The
problem, however, was to show that
    sin(x)^sin(x) < cos(x)^cos(x)
Not the same thing at all.
--------------------------------------------------------------------------------
Carl J Lydick | INTERnet: [e-mail] | NSI/HEPnet: SOL1::CARL

Disclaimer:  Hey, I understand VAXen and VMS.  That's what I get paid for.  My
understanding of astronomy is purely at the amateur level (or below).  So
unless what I'm saying is directly related to VAX/VMS, don't hold me or my
organization responsible for it.  If it IS related to VAX/VMS, you can try to
hold me responsible for it, but my organization had nothing to do with it.