**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.

\(\therefore\text{ Proved}\)

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## Comments

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

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

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

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

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