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rational
 one year ago
solve \(\sin^5x + \cos^5x=1\)
\(x\) is in first quadrant
rational
 one year ago
solve \(\sin^5x + \cos^5x=1\) \(x\) is in first quadrant

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rational
 one year ago
Best ResponseYou've already chosen the best response.2Easy 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
 one year ago
Best ResponseYou've already chosen the best response.0looks cool http://www.wolframalpha.com/input/?i=x^2%2By^2%3D1%2Cx^5%2By^5%3D1

rational
 one year ago
Best ResponseYou've already chosen the best response.2that looks cool indeed xD i know it boils down to \(\sin^2x+\cos^2x=1\) somehow

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that is your only real condition right?

rational
 one year ago
Best ResponseYou've already chosen the best response.2if you mean x is real, yes..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you are solving \[x^5+y^5=1\] given that \[x^2+y^2=1\]

rational
 one year ago
Best ResponseYou've already chosen the best response.2Ahh right! both problems are equivalent

rational
 one year ago
Best ResponseYou've already chosen the best response.2and i feel equally hard/easy

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah and i wouldn't bet more than $7 that there is not some more snappy approach

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0maybe just graph \(x^5+y^5=1\)

rational
 one year ago
Best ResponseYou've already chosen the best response.2I 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
 one year ago
Best ResponseYou've already chosen the best response.2Wow! that does it! thnks @satellite73

rational
 one year ago
Best ResponseYou've already chosen the best response.2sure 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
 one year ago
Best ResponseYou've already chosen the best response.2its not stupid, i think its an alternative approach...

rational
 one year ago
Best ResponseYou've already chosen the best response.2Here is the complete proof : for \(x\in(0,1)\) we have \(\sin^5x\lt \sin^2x\) and \(\cos^5x\lt \cos^2x\). that implies \(\sin^5x+\cos^5x \lt \sin^2x+\cos^2x=1\) \(\blacksquare\)
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