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

Question

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

rational · Answer

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 · Answer

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

rational · Answer

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

anonymous · Answer

that is your only real condition right?

rational · Answer

if you mean x is real, yes..

anonymous · Answer

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

rational · Answer

Ahh right! both problems are equivalent

rational · Answer

and i feel equally hard/easy

anonymous · Answer

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

anonymous · Answer

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

rational · Answer

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 · Answer

Wow! that does it! thnks @satellite73

rational · Answer

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 · Answer

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

rational · Answer

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 · Answer

that solves it