rational
  • rational
solve \(\sin^5x + \cos^5x=1\) \(x\) is in first quadrant
Trigonometry
jamiebookeater
  • jamiebookeater
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rational
  • rational
Easy to see that \(x=0,\pi/2\) are trivial solutions and wolfram gives http://www.wolframalpha.com/input/?i=solve+sin%5E5x%2Bcos%5E5x%3D1%2C+0%3Cx%3Cpi%2F2 idk how to prove that solutions don't exist for \(0\lt x\lt \pi/2\) ..
anonymous
  • anonymous
looks cool http://www.wolframalpha.com/input/?i=x^2%2By^2%3D1%2Cx^5%2By^5%3D1
rational
  • rational
that looks cool indeed xD i know it boils down to \(\sin^2x+\cos^2x=1\) somehow

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anonymous
  • anonymous
that is your only real condition right?
rational
  • rational
if you mean x is real, yes..
anonymous
  • anonymous
you are solving \[x^5+y^5=1\] given that \[x^2+y^2=1\]
rational
  • rational
Ahh right! both problems are equivalent
rational
  • rational
and i feel equally hard/easy
anonymous
  • anonymous
yeah and i wouldn't bet more than $7 that there is not some more snappy approach
anonymous
  • anonymous
maybe just graph \(x^5+y^5=1\)
rational
  • rational
I think so... is it sufficient if we could show that \(x^5+y^5\lt x^2+y^2\) for \(x,y\in (0,1)\)
rational
  • rational
Wow! that does it! thnks @satellite73
rational
  • rational
sure calculus can solve everything :) it is easy to notice that \(a^n \lt a^m\) for \(a\in (0, 1)\) and \(n\gt m\)
rational
  • rational
its not stupid, i think its an alternative approach...
rational
  • rational
Here is the complete proof : for \(x\in(0,1)\) we have \(\sin^5x\lt \sin^2x\) and \(\cos^5x\lt \cos^2x\). that implies \(\sin^5x+\cos^5x \lt \sin^2x+\cos^2x=1\) \(\blacksquare\)
ikram002p
  • ikram002p
that solves it

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