# Factoring Trigonometric Expression.

#### anemone

##### MHB POTW Director
Staff member
Decompose the expression below into real factors:

$$\displaystyle 1-\sin^5 x-\cos^5 x$$

#### chisigma

##### Well-known member
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
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) = $$(1 - \sin x)(1 + \sin x + \sin^2 x + \sin^3 x + \sin^4 x) - \cos^3 x(1 - \sin x)(1 + \sin x) Similarly for 1 - \cos x by 1 - \cos^5x - \sin^3 x(1 - \cos^2 x) but have not been able to get both of them out. The graph suggests that both are there as there are only two zeros on [0,2\pi]. One at x = 0 and the other at x=\pi/2. Last edited: #### Klaas van Aarsen ##### MHB Seeker Staff member If we write z = e^{ix}, the expression becomes$$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.$$So we can further simplify it, which I will do in a later post... Either way, the important factors for the roots are (\sin \frac x 2) and (\cos \frac x 2 - \sin \frac x 2). Last edited: #### Jester ##### Well-known member MHB Math Helper OK, I got the other term. Here's what I have so far. (\sin x -1)(\cos x -1)(3 + 3 \sin x + 3 \cos x + 2 \sin x \cos x - \sin^3 x-\cos^3 x). Next is to show that the 3rd term is always positive. #### Jester ##### Well-known member MHB Math Helper OK, I think I got the rest of this. The third term can be written as \dfrac{1}{2}\left(\sin x + \cos x + \dfrac{2}{3}\right)^3 + \dfrac{5}{6}\left(\sin x +\cos x + \dfrac{2}{3}\right) + \dfrac{35}{27}. So this term is cubic in the variable \sin x + \cos x + \dfrac{2}{3} which is increasing so is minimum would be at the left endpoint of the interval of interest. Now the trig terms can be written as \sin x + \cos x + \dfrac{2}{3} = \sqrt{2} \sin \left( x + \dfrac{\pi}{4}\right) + \dfrac{2}{3} from which we can deduce that the interval of interest is \left[\dfrac{2}{3} - \sqrt{2}, \dfrac{2}{3} + \sqrt{2}\right] Substituting the left endpoint into the cubic shows it is positive, thus giving that 3 + 3 \sin x + 3 \cos x + 2 \sin x \cos x - \sin^3 x - \cos^3 x > 0 #### anemone ##### MHB POTW Director Staff member A possible procedure is to use the identities... \displaystyle \sin x = \frac{2\ t}{1+ t^{2}} \displaystyle \cos x = \frac{1-t^{2}}{1+ t^{2}} (1) ... where \displaystyle t = \tan \frac{x}{2}. In this case we obtain... \displaystyle 1 -\sin^{5} x - \cos^{5} x = \frac{(1+t^{2})^{5} - (1 - t^{2})^{5} - 32\ t^{5}}{(1+t^{2})^{5}} = \frac{2\ t^{2}\ (t-1)^{2}\ (t^{6} + 2\ t^{5} + 3\ t^{4} + 4\ t^{3} + 15\ t^{2} + 10\ t + 5)}{(1+t^{2})^{5}} (2) The problem now is, of course, to pass from the factorization in t to the factorization in x... Kind regards \chi \sigma If we write z = e^{ix}, the expression becomes$$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. 