🚀 go-pugleaf

In article <1992Nov22.030801.4773@dartvax.dartmouth.edu> J.Theodore.Schuerzinger@dartmouth.edu (J. Theodore Schuerzinger) writes:
>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
          ^^^^^^                        ^^^^^^^^
Where did you get (cos x)^2 and (sin x)^2 from ?  You need to
show that (cos x)^(cos x) [which is not the same as (cos x)^2]
is greater than (sin x)^(sin x).

>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'."

Consider the following approach :

The function f(x) = x^(1/sqrt(1-x^2)) is monotonically increasing for
0 < x < 1, easily verified by taking the derivative. 
Since 0 < sin x < cos x < 1 for 0 < x < pi/4, f(sin x) < f(cos x).
But f(sin x) = (sin x)^(1/cos x) and f(cos x) = (cos x)^(1/sin x).
Raising both sides to the power (cos x.sin x), we get the desired
result.

Kanad  Chakraborty