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- Feb 14, 2012

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Decompose the expression below into real factors:

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

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

- Thread starter anemone
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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,894

Decompose the expression below into real factors:

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

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

- 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

$\chi$ $\sigma$

- 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 $x=\pi/2$.

$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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- Mar 5, 2012

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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 $(\cos \frac x 2 - \sin \frac x 2)$.

$$1−\left(\frac 1 {2i}(z-z^{-1})\right)^5−\left(\frac 1 2(z+z^{-1})\right)^5$$

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 $(\cos \frac x 2 - \sin \frac x 2)$.

Last edited:

- Jan 26, 2012

- 183

$(\sin x -1)(\cos x -1)(3 + 3 \sin x + 3 \cos x + 2 \sin x \cos x - \sin^3 x-\cos^3 x)$.

Next is to show that the 3rd term is always positive.

- Jan 26, 2012

- 183

$\dfrac{1}{2}\left(\sin x + \cos x + \dfrac{2}{3}\right)^3 + \dfrac{5}{6}\left(\sin x +\cos x + \dfrac{2}{3}\right) + \dfrac{35}{27}$.

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

- Feb 14, 2012

- 3,894

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 $(\cos \frac x 2 - \sin \frac x 2)$.

And

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

$(\sin x -1)(\cos x -1)(3 + 3 \sin x + 3 \cos x + 2 \sin x \cos x - \sin^3 x-\cos^3 x)$.

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 \(\displaystyle \sin x+\cos x+\frac{2}{3}\).

$\dfrac{1}{2}\left(\sin x + \cos x + \dfrac{2}{3}\right)^3 + \dfrac{5}{6}\left(\sin x +\cos x + \dfrac{2}{3}\right) + \dfrac{35}{27}$.

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

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

Thanks for the posts and insights,

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