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- Feb 14, 2012
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Decompose the expression below into real factors:
\(\displaystyle 1-\sin^5 x-\cos^5 x\)
\(\displaystyle 1-\sin^5 x-\cos^5 x\)
Decompose the expression below into real factors:
\(\displaystyle 1-\sin^5 x-\cos^5 x\)
A possible procedure is to use the identities...
$\displaystyle \sin x = \frac{2\ t}{1+ t^{2}}$
$\displaystyle \cos x = \frac{1-t^{2}}{1+ t^{2}}$ (1)
... where $\displaystyle t = \tan \frac{x}{2}$. In this case we obtain...
$\displaystyle 1 -\sin^{5} x - \cos^{5} x = \frac{(1+t^{2})^{5} - (1 - t^{2})^{5} - 32\ t^{5}}{(1+t^{2})^{5}} = \frac{2\ t^{2}\ (t-1)^{2}\ (t^{6} + 2\ t^{5} + 3\ t^{4} + 4\ t^{3} + 15\ t^{2} + 10\ t + 5)}{(1+t^{2})^{5}}$ (2)
The problem now is, of course, to pass from the factorization in t to the factorization in x...
Kind regards
$\chi$ $\sigma$
Thanks to both of you for showing that the factorization could also be done by using those two different useful trigonometric substitutions. I want to thank to both of you too for taking the time to participating to this challenge problem.If we write $z = e^{ix}$, the expression becomes
$$1−\left(\frac 1 {2i}(z-z^{-1})\right)^5−\left(\frac 1 2(z+z^{-1})\right)^5$$
According to Wolfram, this is equal to
$$\frac{(\frac 1{32}+\frac i{32}) (z-1)^2 (z-i)^2 (1+(2-2 i) z-9 i z^2-(12+12 i) z^3-9 z^4-(2-2 i) z^5+i z^6)}{z^5}$$
As an observation, if we set it to zero, we find the roots $z=1$ and $z=i$ (each of multiplicity 2) that translate to $x=0 \pmod{2\pi}$ and $x=\frac \pi 2 \pmod{2\pi}$.
We can rewrite the expression as
$$\frac 1{32}(1+i) (z^{\frac 1 2}-z^{-\frac 1 2})^2 (z^{\frac 1 2}-iz^{-\frac 1 2})^2 (z^{-3}+(2-2 i) z^{-2}-9 i z^{-1}-12(1+i) -9 z-(2-2 i) z^2+i z^3)$$
$$\frac 1{32}(1+i) (z^{\frac 1 2}-z^{-\frac 1 2})^2 (z^{\frac 1 2}-iz^{-\frac 1 2})^2 (i (z^3 - iz^{-3})-2(1-i)(z^2 - z^{-2})-9 (z+i z^{-1})-12(1+i))$$
We can verify that
$$\left\{ \begin{aligned}
z+z^{-1}&=2\cos x \\
z-z^{-1}&=2i\sin x \\
z-i z^{-1}&=(1-i)(\cos x - \sin x) \\
z+iz^{-1}&=(1+i)(\cos x + \sin x)
\end{aligned}\right.$$
So we can further simplify it, which I will do in a later post...
Either way, the important factors for the roots are $(\sin \frac x 2)$ and $(\cos \frac x 2 - \sin \frac x 2)$.
This is exactly what I did to get the first two real factors andOK, I got the other term. Here's what I have so far.
$(\sin x -1)(\cos x -1)(3 + 3 \sin x + 3 \cos x + 2 \sin x \cos x - \sin^3 x-\cos^3 x)$.
Next is to show that the 3rd term is always positive.
I am impressed with how easy the desired result you obtained (i.e. to prove the third factor is always positive for all real \(\displaystyle x\)) by setting up the third factor in terms of \(\displaystyle \sin x+\cos x+\frac{2}{3}\).OK, I think I got the rest of this. The third term can be written as
$\dfrac{1}{2}\left(\sin x + \cos x + \dfrac{2}{3}\right)^3 + \dfrac{5}{6}\left(\sin x +\cos x + \dfrac{2}{3}\right) + \dfrac{35}{27}$.
So this term is cubic in the variable $\sin x + \cos x + \dfrac{2}{3}$ which is increasing so is minimum would be at the left endpoint of the interval of interest. Now the trig terms can be written as
$\sin x + \cos x + \dfrac{2}{3} = \sqrt{2} \sin \left( x + \dfrac{\pi}{4}\right) + \dfrac{2}{3}$
from which we can deduce that the interval of interest is
$\left[\dfrac{2}{3} - \sqrt{2}, \dfrac{2}{3} + \sqrt{2}\right]$
Substituting the left endpoint into the cubic shows it is positive, thus giving that
$3 + 3 \sin x + 3 \cos x + 2 \sin x \cos x - \sin^3 x - \cos^3 x > 0$