Chebyshev Polynomial Practice: Product

Let $T_n$ be the nth Chebyshev Polynomial where $T_n(\cos(x))=\cos(nx)$

Rearrange the equation to

$\sec(20)\sec(100)\sec(140)$

Note that we can do this because we changes two signs using the identity $\cos(x)=-\cos(180-x)$ .

We are looking at the CP $~~T_3(x)=\frac{1}{2}$ . Since $\cos(60)=\cos(300)=\cos(420)=\frac{1}{2}$

$T_{odd}(x)$ has no constant term so we have

$T_3(x)-\frac{1}{2}=4x^3-3x-\frac{1}{2}=0$

By vietas, the product of the roots is

$-\cfrac{-\frac{1}{2}}{4}=\dfrac{1}{8}$

Since $\sec(x)$ is the reciprocal of $\cos(x)$ , the product of the roots of $\sec(x)=8$

$\sec20\sec40\sec80=\frac{1}{\cos20\cos40\cos80}$ $=\frac{1}{\frac{1}{2}(\cos20+\cos60)\cos80}=\frac{1}{\frac{1}{2}\cos20\cos80+\frac{1}{4}\cos80}$ $=\frac{1}{\frac{1}{2}(\frac{1}{2}(\cos60+\cos100))+\frac{1}{4}\cos80}=\frac{1}{\frac{1}{8}+\frac{1}{4}(\cos100+\cos80)}$ $=\frac{1}{\frac{1}{8}+\frac{1}{4}(2\cos90\cos10)}=\frac{1}{\frac{1}{8}+\frac{1}{4}(0)}=\frac{1}{\frac{1}{8}}=\bf 8 \space\space\space\Box$

sec(20)sec(40)sec(80)= 4sec3 20 =4 sec60 =4 2=8