Chebyshev roots

Chebyshev polynomials often appear in HSC Extension topics where trigonometric identities meet polynomial equations. By using the natural substitution (x = \cos\theta), we turn the polynomial (T_5(x)) into the simple trigonometric equation (\cos(5\theta) = 0). This approach lets you bypass factoring a quintic and instead solve directly for the five roots on ([-1,1]) — a powerful trick for your problem-solving toolkit.

Problem Statement

The first-kind Chebyshev polynomial of degree $5$ is

$$T_5(x)=16x^5-20x^3+5x.$$

Use the identity $T_5(\cos\theta)=\cos(5\theta)$ to find all roots of $T_5(x)$ in the interval $[-1,1]$.

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Hints

Set $\cos(5\theta)=0$, then convert the resulting angles back to $x=\cos\theta$.

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Solutions

Let $x=\cos\theta$. We need

$$\cos(5\theta)=0.$$

To satisfy this, $5\theta$ must be an odd multiple of $\frac{\pi}{2}$, because the cosine of any odd multiple of $\frac{\pi}{2}$ is zero. Thus we set

$$5\theta=\frac{(2k+1)\pi}{2},\qquad k=0,1,2,3,4.$$

Dividing through by $5$ gives the possible angles

$$\theta=\frac{(2k+1)\pi}{10},$$

and the five roots are obtained by converting back to $x$ with $x=\cos\theta$:

$$x=\cos\frac{\pi}{10},\ \cos\frac{3\pi}{10},\ \cos\frac{5\pi}{10},\ \cos\frac{7\pi}{10},\ \cos\frac{9\pi}{10}.$$

These are the only distinct roots because $k=0,1,2,3,4$ covers all angles $\theta$ in $[0,\pi]$ that produce different values of $\cos\theta$, and the polynomial is degree $5$.

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Takeaways

• Roots of $T_n(x)$ on $[-1,1]$ come from zeros of $\cos(n\theta)$.
• The substitution $x=\cos\theta$ is natural because $\cos\theta$ covers $[-1,1]$.
• Degree $5$ gives five roots, which is a useful check.

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Further Readings

HSC Combinatorics, HSC Mechanics, HSC Proofs, HSC Trigonometry