We begin by recalling the definition of an even and odd function.
An even function has the property that
\[f(-x) = f(x)\]
In other words, this function is symmetric across the y-axis.
An odd function has the property that
\[f(-x) = -f(x)\]
In other words, this function is rotationally symmetric about the origin. That is, in \(\mathbb{R}^2\),
the function can be rotated \(180^{\circ}\) and remain unchanged.
Now we will consider the trigonometric Fourier Series of a function on the circle \((-\pi, \pi]\).
\[
f(\theta) = a_0 + \sum_{n = 1}^{\infty}\left(a_n\cos\left(n\theta\right) + b_n\sin\left(n\theta\right)\right)
\]
\[
a_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(\theta)\,d\theta
\]
\[
a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(\theta)\cos(n\theta)\,d\theta
\]
\[
b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(\theta)\sin(n\theta)\,d\theta
\]
I will show that a function that is even will be comprised of only cosine waves, as cosine is an even function.
Similarly, I will show that an odd function will be comprised of only sine waves, as sine is an odd function.
Even \(f\)
Let \(f(\theta)\) be an even function, defined on \((-\pi, \pi]\). Then we know that \(f(-\theta) = f(\theta)\).
Now let's consider its Fourier Series representation.
\[
f(\theta) = a_0 + \sum_{n = 1}^{\infty}a_n\cos(n\theta) + b_n\sin(n\theta)
\]
Then by the property of \(f\) being even, we know that \(f(\theta) = f(-\theta)\), so let's analyze the Fourier coefficients.
\[
a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(\theta)\cos(n\theta)\,d\theta
= \frac{1}{\pi}\int_{-\pi}^{\pi}f(-\theta)\cos(-n\theta)\,d\theta
= \frac{2}{\pi}\int_{0}^{\pi}f(\theta)\cos(n\theta)\,d\theta
\]
Above follows directly from substituting in \(f(\theta) = f(-\theta)\) and \(\cos(n\theta) = \cos(-n\theta)\), since both are even functions.
\[
b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(\theta)\sin(n\theta)\,d\theta
= \frac{1}{\pi}\left(\int_{0}^{\pi}f(\theta)\sin(n\theta)\,d\theta
+ \int_{-\pi}^{0}f(\theta)\sin(n\theta)\,d\theta\right)
\]
Note that
\[
\int_{-\pi}^{0}f(\theta)\sin(n\theta)\,d\theta
= \int_{0}^{\pi}f(-\theta)\sin(-n\theta)\,d\theta
= -\int_{0}^{\pi}f(\theta)\sin(n\theta)\,d\theta
\]
By the property that \(f\) is even, and \(\sin\) is odd. Substituting this in, we get
\[
\frac{1}{\pi}\left(\int_{0}^{\pi}f(\theta)\sin(n\theta)\,d\theta
- \int_{0}^{\pi}f(\theta)\sin(n\theta)\,d\theta\right) = 0
\]
Therefore, we see that
\[
\forall n \in \mathbb{N} \text{ and } \forall \theta \in (-\pi, \pi] \quad b_n = 0
\]
Finally, our Fourier Series of our even function is simply
\[
f(\theta) = a_0 + \sum_{n=1}^{\infty}a_n\cos(n\theta)
\]
Odd \(f\)
Now we will assume \(f\) is odd and show that its Fourier series is only comprised of sine waves. So, we have that
\[
f(-\theta) = -f(\theta)
\]
We will consider the trigonometric Fourier Series of a function on the circle \((-\pi, \pi]\).
\[
f(\theta) = a_0 + \sum_{n = 1}^{\infty}\left(a_n\cos\left(n\theta\right) + b_n\sin\left(n\theta\right)\right)
\]
\[
a_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(\theta)\,d\theta
\]
\[
a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(\theta)\cos(n\theta)\,d\theta
\]
\[
b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(\theta)\sin(n\theta)\,d\theta
\]
Let's analyze \(a_n\) more closely,
\[
a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(\theta)\cos(n\theta)\,d\theta
= \frac{1}{\pi}\left(\int_{0}^{\pi}f(\theta)\cos(n\theta)\,d\theta
+ \int_{-\pi}^{0}f(\theta)\cos(n\theta)\,d\theta\right)
\]
Like before,
\[
\int_{-\pi}^{0}f(\theta)\cos(n\theta)\,d\theta
= \int_{0}^{\pi}f(-\theta)\cos(-n\theta)\,d\theta
\]
Plugging that in above we have,
\[
a_n = \frac{1}{\pi}\left(\int_{0}^{\pi}f(\theta)\cos(n\theta)\,d\theta
+ \int_{0}^{\pi}f(-\theta)\cos(-n\theta)\,d\theta\right)
\]
By the property that \(f(-\theta) = -f(\theta)\) and \(\cos(-n\theta) = \cos(n\theta)\), we have
\[
\int_{0}^{\pi}f(-\theta)\cos(-n\theta)\,d\theta = -\int_{0}^{\pi}f(\theta)\cos(n\theta)\,d\theta
\]
Making this final substitution, we have
\[
a_n = \frac{1}{\pi}\left(\int_{0}^{\pi}f(\theta)\cos(n\theta)\,d\theta
- \int_{0}^{\pi}f(\theta)\cos(n\theta)\,d\theta\right) = 0
\]
Therefore, we have shown that
\[
a_n = 0
\]
For any odd function on the circle \((-\pi, \pi]\). Finally,
\[
f(\theta) = \sum_{n = 1}^{\infty}b_n\sin\left(n\theta\right)
\]
We see that any odd function is comprised of only sine waves in its Fourier decomposition.