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

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1. 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 $$0\lt x\lt \pi/2$$ ..

2. anonymous
3. rational

that looks cool indeed xD i know it boils down to $$\sin^2x+\cos^2x=1$$ somehow

4. anonymous

that is your only real condition right?

5. rational

if you mean x is real, yes..

6. anonymous

you are solving $x^5+y^5=1$ given that $x^2+y^2=1$

7. rational

Ahh right! both problems are equivalent

8. rational

and i feel equally hard/easy

9. anonymous

yeah and i wouldn't bet more than \$7 that there is not some more snappy approach

10. anonymous

maybe just graph $$x^5+y^5=1$$

11. rational

I think so... is it sufficient if we could show that $$x^5+y^5\lt x^2+y^2$$ for $$x,y\in (0,1)$$

12. rational

Wow! that does it! thnks @satellite73

13. rational

sure calculus can solve everything :) it is easy to notice that $$a^n \lt a^m$$ for $$a\in (0, 1)$$ and $$n\gt m$$

14. rational

its not stupid, i think its an alternative approach...

15. 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 $$\sin^5x+\cos^5x \lt \sin^2x+\cos^2x=1$$ $$\blacksquare$$

16. ikram002p

that solves it