There’s more juice left in the lemon we’ve been squeezing lately.
A few days ago I first brought up the equation
which holds because both sides equal exp(inθ).
Then a couple days ago I concluded a blog post by noting that by taking the real part of this equation and replacing sin²θ with 1 – cos²θ one could express cos nθ as a polynomial in cos θ,
and in fact this polynomial is the nth Chebyshev polynomial Tn since these polynomials satisfy
Now in this post I’d like to prove a relationship between Chebyshev polynomials and sines starting with the same raw material. The relationship between Chebyshev polynomials and cosines is well known, even a matter of definition depending on where you start, but the connection to sines is less well known.
Let’s go back to the equation at the top of the post, replace n with 2n + 1, and take the imaginary part of both sides. The odd terms of the sum contribute to the imaginary part, so we sum over 2ℓ+ 1.
Here we did a change of variables k = n – ℓ.
The final expression is the expression we began with, only evaluated at sin θ instead of cos θ. That is,
So for all n we have
and for odd n we also have
The sign is positive when n is congruent to 1 mod 4 and negative when n is congruent to 3 mod 4.