Painevg

## Answered question

2021-12-28

Number of solutions of ${\mathrm{cos}}^{5}x+{\mathrm{cos}}^{5}\left(x+\frac{2\pi }{3}\right)+{\mathrm{cos}}^{5}\left(x+\frac{4\pi }{3}\right)=0$

### Answer & Explanation

amarantha41

Beginner2021-12-29Added 38 answers

Let
$f\left(x\right)={\mathrm{cos}}^{5}x+{\mathrm{cos}}^{5}\left(x+\frac{2\pi }{3}\right)+{\mathrm{cos}}^{5}\left(x+\frac{4\pi }{3}\right)$
Because cosine has period $2\pi$, we easily see that f(x) has period $\frac{2\pi }{3}$. It is well behaved (continuously differentiable), so it can be expanded into a Fourier series. The function f(x) is even, f(-x)=f(x) for all x, so the expansion only contains cosine terms. As you observed, it is a quintic polynomial in cosx, so
$f\left(x\right)=\sum _{k=0}^{5}{a}_{k}\mathrm{cos}kx$
for some constants ${a}_{k}\in \mathbb{R}$. Period $\frac{2\pi }{3}$ implies that ${a}_{k}=0$ unless k is divisible by three. At this point we know that
$f\left(x\right)={a}_{0}+{a}_{3}\mathrm{cos}3x$
for some constants ${a}_{0},{a}_{3}$. We have $f\left(x+\pi \right)=-f\left(x\right)$ for all x implying that ${a}_{0}=0$ (you can see this also by observing that only odd powers of $\mathrm{cos}x$ occur in that quintic, or by observing that ${\int }_{0}^{2\pi }f\left(x\right)dx=0\right)$
Last but not least
${a}_{3}=f\left(0\right)=1+{\left(\frac{-1}{2}\right)}^{5}+{\left(\frac{-1}{2}\right)}^{5}=\frac{15}{16}$
So
$f\left(x\right)=\frac{15}{16}\mathrm{cos}3x$

Alex Sheppard

Beginner2021-12-30Added 36 answers

Use $\mathrm{cos}\left(A+B\right)$ formula to show for a natural number n,
$\mathrm{cos}nx+\mathrm{cos}\left(nx+\frac{2n\pi }{3}\right)+\mathrm{cos}\left(nx+\frac{4n\pi }{3}\right)=\left\{\begin{array}{cc}0& \text{when n is coprime to 3}\\ 3\mathrm{cos}nx& \text{when n is a multiple of 3}\end{array}$
Combining this with the identity from other answer,
${\mathrm{cos}}^{5}\theta =\frac{1}{16}\mathrm{cos}5\theta +\frac{5}{16}\mathrm{cos}3\theta +\frac{10}{16}\mathrm{cos}\theta$
it is easy to see that first and third terms, n=1,5, should sum to zero, while middle term remains to give
${\mathrm{cos}}^{5}x+{\mathrm{cos}}^{5}\left(x+\frac{2\pi }{3}\right)+{\mathrm{cos}}^{5}\left(x+\frac{4\pi }{3}\right)=\frac{15}{16}\mathrm{cos}3x$

nick1337

Expert2022-01-08Added 777 answers

Hint:
${\mathrm{cos}}^{5}\theta =\frac{10\mathrm{cos}\left(\theta \right)+5\mathrm{cos}\left(3\theta \right)+\mathrm{cos}\left(5\theta \right)}{16}$
because
${\mathrm{cos}}^{5}x=\frac{1}{32}\left({e}^{ix}+{e}^{-ix}{\right)}^{5}=\frac{{e}^{5ix}+{e}^{-5ix}+5\left({e}^{3ix}+{e}^{-3ix}\right)+10\left({e}^{ix}+{e}^{-ix}\right)}{32}$
Note that there will be infinitely many solutions as pointed out in comments

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