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Factoring Trigonometric Expression.

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,678
Decompose the expression below into real factors:

\(\displaystyle 1-\sin^5 x-\cos^5 x\)
 

chisigma

Well-known member
Feb 13, 2012
1,704
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


$\chi$ $\sigma$
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
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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Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
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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Jester

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MHB Math Helper
Jan 26, 2012
183
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.
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
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

$3 + 3 \sin x + 3 \cos x + 2 \sin x \cos x - \sin^3 x - \cos^3 x > 0$
 
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  • #7

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,678
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


$\chi$ $\sigma$

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.:)

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

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 \(\displaystyle x\)) by setting up the third factor in terms of \(\displaystyle \sin x+\cos x+\frac{2}{3}\).

Thanks for the posts and insights, Jester!:)

P.S. The way that I proved the third factor is always positive for all real \(\displaystyle x\) is by the graphing method.:eek: