Expressing \(\cos(n\theta)\) and \(\sin(n\theta)\) as polynomials of \(\sin \theta\) and \(\cos \theta\).
Let&nbsp;\(n\) be a positive integer. On one hand, by the De Moivre's Theorem,
\[(\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta).\]
On the other hand,
\[(\cos \theta + i \sin \theta)^n = \sum_{k=0}^{n} \binom{n}{k} \cos^{n-k} \theta (i \sin \theta)^k.\]
Separating the real and imaginary parts and comparing with the real and imaginary parts of \((\cos \theta + i \sin \theta)^n\), we have that
\[\cos(n\theta) = \cos^n \theta - \binom{n}{2} \cos^{n-2} \theta \sin^2 \theta + \binom{n}{4} \cos^{n-4} \theta \sin^4 \theta - \dots\]
and
\[\sin(n\theta) = \binom{n}{1} \cos^{n-1} \theta \sin \theta - \binom{n}{3} \cos^{n-3} \theta \sin^3 \theta - \dots\]
Because all powers of \(\cos \theta\) and \(\sin \theta\)'s are even in the expression of \(\cos(n\theta)\), we may convert all the \(\sin^2 \theta\) into \(1 - \cos^2 \theta\), and simplify \(\cos(n\theta)\) as a polynomial \(T_n\) of \(\cos \theta\), namely,
\[T_n(\cos \theta) = \cos(n\theta).\]
These polynomials \(\{T_n\}\) are called the Chebyshev polynomials of the first kind.
For the expression of \(\sin(n\theta)\), we may factor out \(\sin \theta\) and see that
\[\sin(n\theta) = \left[ \binom{n}{1} \cos^{n-1} \theta - \binom{n}{3} \cos^{n-3} \theta \sin^2 \theta - \dots \right] (\sin \theta).\]
Thus, by convert all the occurrences of \(\sin^2 \theta\) into \(1 - \cos^2 \theta\) we again can express
\[\sin(n\theta) = U_n(\cos \theta) \cdot \sin \theta,\]
where&nbsp;\(\{U_n\}\) are called the Chebyshev polynomials of the second kind.

**Expressing \(\cos(n\theta)\) and \(\sin(n\theta)\) as polynomials of \(\sin \theta\) and \(\cos \theta\).**

Let \(n\) be a positive integer. On one hand, by the De Moivre's Theorem,

\[(\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta).\]

\[(\cos \theta + i \sin \theta)^n = \sum_{k=0}^{n} \binom{n}{k} \cos^{n-k} \theta (i \sin \theta)^k.\]

Separating the real and imaginary parts and comparing with the real and imaginary parts of \((\cos \theta + i \sin \theta)^n\), we have that

\[\cos(n\theta) = \cos^n \theta - \binom{n}{2} \cos^{n-2} \theta \sin^2 \theta + \binom{n}{4} \cos^{n-4} \theta \sin^4 \theta - \dots\]

\[\sin(n\theta) = \binom{n}{1} \cos^{n-1} \theta \sin \theta - \binom{n}{3} \cos^{n-3} \theta \sin^3 \theta - \dots\]

Because all powers of \(\cos \theta\) and \(\sin \theta\)'s are even in the expression of \(\cos(n\theta)\), we may convert all the \(\sin^2 \theta\) into \(1 - \cos^2 \theta\), and simplify \(\cos(n\theta)\) as a polynomial \(T_n\) of \(\cos \theta\), namely,

These polynomials \(\{T_n\}\) are called the **Chebyshev polynomials of the first kind**.

For the expression of \(\sin(n\theta)\), we may factor out \(\sin \theta\) and see that

\[\sin(n\theta) = \left[ \binom{n}{1} \cos^{n-1} \theta - \binom{n}{3} \cos^{n-3} \theta \sin^2 \theta - \dots \right] (\sin \theta).\]

Thus, by convert all the occurrences of \(\sin^2 \theta\) into \(1 - \cos^2 \theta\) we again can express

\[\sin(n\theta) = U_n(\cos \theta) \cdot \sin \theta,\]

where \(\{U_n\}\) are called the **Chebyshev polynomials of the second kind**.

Properties from Exponential or Trigonometric form [Out of Syllabus]

Complex Numbers

International Baccalaureate Higher Level Mathematics

Multiple angle formulae

Multiple angle formulae