Proof of a Chebyshev Identity

Reference: Wiki on Chebyshev Polynomials, Proof problem 6.

$\huge T_n(x) = \frac{ ( x - \sqrt{ x^2 - 1} )^n + ( x + \sqrt{ x^2 - 1 } ) ^ n } { 2} .$

Solution: $\text{From the above identity, we get}$ $\large{ T }_{ n }(\cos { \theta )= } \frac { { \left( \cos { \theta -\sqrt { \cos ^{ 2 }{\theta}- 1 } } \right) }^{ n }+{ \left( \cos { \theta +\sqrt { \cos ^{ 2 }{ \theta } -1} } \right) }^{ n } }{ 2 } =\frac { { \left( \cos { \theta -i\sin { \theta } } \right) }^{ n }+{ \left( \cos { \theta +i\sin { \theta } } \right) }^{ n } }{ 2 }$.

$\text{Using}$ De Moivre's Theorem, $\text{we get}$ $\large\frac { { \left( \cos { \theta -i\sin { \theta } } \right) }^{ n }+{ \left( \cos { \theta +i\sin { \theta } } \right) }^{ n } }{ 2 } =\frac { \cos { n\theta -i\sin { \theta +\cos { n\theta +i\sin { n\theta } } } } }{ 2 } =\cos { n\theta }$.

$\text{This turns out to be the definition of the}$ Chebyshev Polynomial of the first kind.

$\therefore\text{ Proved}$

Note by Swapnil Das
3 years, 10 months ago

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Nice work. I have a doubt $\sqrt{1-cos^2(\theta)}=\sin(\theta)$. Can you please explain how you got $i\sin(\theta)$ and how $\sqrt{1+cos^2(\theta)}=i\sin(\theta)$.

- 3 years, 10 months ago

Correct. That's a typo, thanks for pointing out!

- 3 years, 10 months ago

Yup, I don't understand how Swapnil wrote $\sqrt{1-\cos^2 \theta}$ and $\sqrt{1+\cos^2 \theta}$

- 3 years, 10 months ago

Bro, I am Swapnil. Is it fine now? (It was a typo)

- 3 years, 10 months ago

Yep, it is correct now. Sorry, Swaplin was a typo, I meant to type Swapnil :P :P :P

- 3 years, 10 months ago