Are there any proofs of this equality online? I'm just looking for something very simply that I can self-verify. My textbook uses the result without a proof, and I want to see what a proof would look like here.
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3You can use induction: express $\cos((n+1)x)$ in terms of $\cos nx$ and $\cos x$. Also, look up "Chebyshev polynomials" and "cosine". – Nov 12 '14 at 00:28
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\begin{align} e^{inx} &= (\cos x + i\sin x)^n \\&= \sum_{k=0}^n \binom nk\cos^{n-k}(x) i^k \sin^k (x) \\&= \sum_{k=0}^{\lfloor n/2\rfloor} \binom n{2k}\cos^{n-2k}(x) i^{2k} \sin^{2k} (x) + \sum_{k=0}^{\lfloor (n-1)/2\rfloor} \binom n{2k+1}\cos^{n-2k-1}(x) i^{2k+1} \sin^{2k+1} (x) \end{align}
Take the real part of both sides and you get $$ \cos nx =\sum_{k=0}^{\lfloor n/2\rfloor} \binom n{2k}\cos^{n-2k}(x) (-1)^k (1-\cos^2x)^k $$
The last expression is clearly a polynomial of degree $n$ in $\cos (x)$.
Alternative:
as $$ \cos ((n+1)x) + \cos ((n-1)x) = 2\cos (x) \cos (nx) $$ you can prove it directly using induction.
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Does $\cos(x)\cos(nx)$ reduce to a square or something? I don't see where the polynomial comes in. – Ashley Tisdle Nov 12 '14 at 00:57
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if $P_k$ is the polynomial for $\cos kx$ for $k$ up to $n$, then this proves that $\cos (n+1)x = 2\cos x \cos nx - \cos(n-1)x$. So that $P_{n+1}$ defined by $P_{n+1}(X) = 2XP_n(X) - P_{n-1}(X)$ is convenient. – mookid Nov 12 '14 at 01:02
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I am not sure which proof you are looking for.
Your polynomial $Q_n$ is called the Chebyshev polynomial of degree $N$.
Probably the wikipedia page (http://en.wikipedia.org/wiki/Chebyshev_polynomials) will help you out in finding what you are looking for.
Andreas H.
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We have \begin{align} \cos((n+1)x) & = \cos(x) \cos(nx) - \sin(x) \sin(nx)\\ & = \cos(x) \cos(nx) + \dfrac{\cos((n+1)x) - \cos((n-1)x)}2 \end{align} Hence, $$2 \cos((n+1)x) = 2\cos(x) \cos(nx) + \cos((n+1)x) - \cos((n-1)x)$$ Therefore, $$\cos((n+1)x) = 2\cos(x) \cos(nx) - \cos((n-1)x)$$ Now use induction to conclude what you want.
Adhvaitha
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