MHB Factoring Trigonometric Expression.

AI Thread Summary
The discussion focuses on factoring the expression \(1 - \sin^5 x - \cos^5 x\) using trigonometric identities and substitutions. Participants suggest using the half-angle tangent substitution and explore the expression's transformation into a polynomial in terms of \(t\). Key factors identified include \((\sin x - 1)\) and \((\cos x - 1)\), with further analysis leading to a cubic term that is shown to be always positive. The conversation emphasizes the importance of these factors in determining the roots of the original expression, confirming that the expression is valid across the specified interval. Overall, the collaborative effort highlights various approaches to factorization and verification of positivity in trigonometric expressions.
anemone
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Decompose the expression below into real factors:

$$1-\sin^5 x-\cos^5 x$$
 
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anemone said:
Decompose the expression below into real factors:

$$1-\sin^5 x-\cos^5 x$$
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 $\chi$ $\sigma$
 
I think it's easy to deduce that $1 - \sin x$ is a factor. If we re-write our expression as

$1 - \sin^5 x - \cos^3x(1-\sin^2 x) = $$(1 - \sin x)(1 + \sin x + \sin^2 x + \sin^3 x + \sin^4 x) - \cos^3 x(1 - \sin x)(1 + \sin x)$

Similarly for $1 - \cos x$ by $1 - \cos^5x - \sin^3 x(1 - \cos^2 x)$ but have not been able to get both of them out. The graph suggests that both are there as there are only two zeros on $[0,2\pi]$. One at $x = 0$ and the other at $x=\pi/2$.
 
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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 $(\cos \frac x 2 - \sin \frac x 2)$.
 
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OK, 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.
 
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

$3 + 3 \sin x + 3 \cos x + 2 \sin x \cos x - \sin^3 x - \cos^3 x > 0$
 
chisigma said:
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 $\chi$ $\sigma$
I like Serena said:
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 $(\cos \frac x 2 - \sin \frac x 2)$.

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.

And I like Serena, I am looking forward to see your next post because I know what you are going to post will benefit the readers for sure.:)

Jester said:
OK, 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.

This is exactly what I did to get the first two real factors and

Jester said:
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

$3 + 3 \sin x + 3 \cos x + 2 \sin x \cos x - \sin^3 x - \cos^3 x > 0$

I am impressed with how easy the desired result you obtained (i.e. to prove the third factor is always positive for all real $$x$$) by setting up the third factor in terms of $$\sin x+\cos x+\frac{2}{3}$$.

Thanks for the posts and insights, Jester!:)

P.S. The way that I proved the third factor is always positive for all real $$x$$ is by the graphing method.:o
 
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