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 Inequality (Posted on 2015-02-11)
Show that (sin x)^(sin x) < (cos x)^(cos x) when 0 < x < pi/4

 No Solution Yet Submitted by Danish Ahmed Khan No Rating

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 Solution Comment 5 of 5 |
Using the power series   loge(1 - t) =  – t - t2/2 – t3/3 ….  and

the identity  cos2x = 1 – sin2x :

loge(cos2x) = -sin2x – sin4x/2 – sin6x/3 -…

Thus     2*loge(c) = -s2 – s4/2 – s6/3 …     (s & c for sinx and cosx)

2c*loge(c) = -sc(s + s3/2 + s5/2 +…)                     (1)

Now, starting with sin2x = 1 – cos2x, the same procedure gives:

2s*loge(s) = -cs(c + c3/2 + c5/3 + …)                    (2)

(2) – (1) now gives:

2[s*loge(s) – c*loge(c)] = sc[(s – c) + (s3 – c3)/2 + (s5 – c5)/3 …

For 0 < x < pi/4, 0 < s < c < 1, so the RHS is clearly negative,

and therefore     s*loge(s) < c*loge(c),

which gives:   loge(ss) < loge(cc) and   ss < cc as required.

I’m hoping someone can find an easier way than this..

 Posted by Harry on 2015-02-15 14:54:25

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