Factor the trigonometric expression:$1-\cos^5x-\sin^5x$

Question

Factor the following expression:

$$1-\cos^5x-\sin^5x$$ I have tried to put $\sin^2x+\cos^2x$ instead of $1$ but no results.

Can anyone help please?

Answer 1

Hint:

If $c^n+s^n=(c^2+s^2)(c^n+s^n)=c^{n+2}+s^{n+2}+c^2s^2(c^{n-2}+s^{n-2})$

$\iff c^{n+2}+s^{n+2}=c^n+s^n-c^2s^2(c^{n-2}+s^{n-2})$

$n=3\implies$

$$c^5+s^5=c^3+s^3-c^2s^2(c+s)=(c+s)^3-3cs(c+s)-c^2s^2(c+s)$$

If $c+s=u,u^2-1=2cs$

Replace $c+s, cs$ in terms of $u$

Answer 2

$$1-\cos^5x-\sin^5x = \sin^2x + \cos^2x - \cos^5x -\sin^5x $$

$$=$$

$$(\sin^2x -\sin^5x) + (\cos^2x - \cos^5x) $$

$$=$$

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

This is the only thing that I see as simple enough to be carried out. (You could replace 1 with $\sin^2x + \cos^2x$ and continue again with 1 factorization)

Wolfram Alpha doesn't give anything simple as well (only some complicated expressions).

Answer 3

The set of points $\{(x,y)\in\mathbb{R}^2:x^5+y^5=1\}$ intersects the set of points $\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}$ only at $(0,1)$ and $(1,0)$. So, if the purpose of factoring that expression is to find its real zeroes, you are just wasting time, because you know in advance that the real zeroes lie at $\left\{0,\frac{\pi}{2}\right\}+2\pi\mathbb{Z}$ only.

Answer 4

$$a^5+b^5= (a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)$$

Hence, $$\cos^5x+\sin^5x= (\cos x+\sin x)(\cos^4 x-\cos^3 x\sin x+\cos^2 x\sin^2x-\cos x\sin^3x+\sin^4x)\\= (\cos x+\sin x)(\cos^4 x+\sin^4x -\sin x\cos x\color{red}{(\cos^2 x+\sin^2 x)}+\cos^2 x\sin^2x)\\=(\cos x+\sin x)(\cos^4 x+\sin^4x -\sin x\cos x+\cos^2 x\sin^2x)$$

But: $$\cos^4 x+\sin^4x =(\cos^2 x+\sin^2x )^2-2\cos^2 x\sin^2x = 1-2\cos^2 x\sin^2x $$

Thus, $$\cos^5x+\sin^5x=(\cos x+\sin x)(1-2\cos^2 x\sin^2x -\sin x\cos x(\cos^2 x+\sin^2 x)+\cos^2 x\sin^2x)\\=(\cos x+\sin x)(1-\cos^2 x\sin^2x -\sin x\cos x)$$

Answer 5

Well, this expression can't be simplified to any simple term, but you can easily reduce the power of the terms.

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

$\implies 1-(\frac{e^{ix}+e^{-ix}}{2})^5 - (\frac{e^{ix}-e^{-ix}}{2i})^5$

$\implies 1-(\frac{a+\frac{1}{a}}{2})^5 + i(\frac{a+\frac{1}{a}}{2})^5\quad $ $ ;\ a=e^{ix} $

From here on, it's just binomial theorem as most of the terms are cancelled out /can be regrouped using DeMoivre's theorem.