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rational

  • one year ago

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

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  1. rational
    • one year ago
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    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 \(0\lt x\lt \pi/2\) ..

  2. anonymous
    • one year ago
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    looks cool http://www.wolframalpha.com/input/?i=x^2%2By^2%3D1%2Cx^5%2By^5%3D1

  3. rational
    • one year ago
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    that looks cool indeed xD i know it boils down to \(\sin^2x+\cos^2x=1\) somehow

  4. anonymous
    • one year ago
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    that is your only real condition right?

  5. rational
    • one year ago
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    if you mean x is real, yes..

  6. anonymous
    • one year ago
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    you are solving \[x^5+y^5=1\] given that \[x^2+y^2=1\]

  7. rational
    • one year ago
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    Ahh right! both problems are equivalent

  8. rational
    • one year ago
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    and i feel equally hard/easy

  9. anonymous
    • one year ago
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    yeah and i wouldn't bet more than $7 that there is not some more snappy approach

  10. anonymous
    • one year ago
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    maybe just graph \(x^5+y^5=1\)

  11. rational
    • one year ago
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    I think so... is it sufficient if we could show that \(x^5+y^5\lt x^2+y^2\) for \(x,y\in (0,1)\)

  12. rational
    • one year ago
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    Wow! that does it! thnks @satellite73

  13. rational
    • one year ago
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    sure calculus can solve everything :) it is easy to notice that \(a^n \lt a^m\) for \(a\in (0, 1)\) and \(n\gt m\)

  14. rational
    • one year ago
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    its not stupid, i think its an alternative approach...

  15. rational
    • one year ago
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    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 \(\sin^5x+\cos^5x \lt \sin^2x+\cos^2x=1\) \(\blacksquare\)

  16. ikram002p
    • one year ago
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    that solves it

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