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Article #2355Re: Trig. inequality
From: J. Theodore Schuerzinger
Date: Sun, 22 Nov 1992 03:08
Date: Sun, 22 Nov 1992 03:08
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1460 bytes
In article <1ekddfINNi96@uwm.edu> radcliff@csd4.csd.uwm.edu (David G Radcliffe) writes: > Show that (sin x)^(sin x) < (cos x)^(cos x) when 0 < x < pi/4. > > -- > David Radcliffe > radcliff@csd4.csd.uwm.edu 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! --Ted Schuerzinger email: .zed@Dartmouth.EDU "I should have realized it would be bad vodka when all the label said was 'Russian Vodka'."
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Path: arc1-ca.pugleaf.net!sparky!uunet!cs.utexas.edu!zaphod.mps.ohio-state.edu!sol.ctr.columbia.edu!destroyer!gumby!yale!hsdndev!dartvax!J.Theodore.Schuerzinger From: J.Theodore.Schuerzinger@dartmouth.edu (J. Theodore Schuerzinger) Newsgroups: rec.puzzles Subject: Re: Trig. inequality Message-ID: <1992Nov22.030801.4773@dartvax.dartmouth.edu> Date: 22 Nov 92 03:08:01 GMT References: <1ekddfINNi96@uwm.edu> Sender: news@dartvax.dartmouth.edu (The News Manager) Organization: Dartmouth College, Hanover, NH Lines: 23 X-Posted-From: InterNews1.0a5@newshost.dartmouth.edu