Are sin functions required for representing a function using a Fourier series, or are cos functions on their own sufficient?

Question

The Fourier series $$f(t) = d + \sum_{n=-1}^\infty a_n \cos(nt) + b_n \sin(nt),$$

can be written in the form

$$\sum_{n=-\infty}^\infty c_n e^{int},$$

where

$c_n = \begin{cases} d \quad \text{for} \ n = 0 \\ (a - ib_n)/2, \quad \text{for} \ n \ \text{positive} \\ (a_{-n} + ib_{-n})/2, \quad \text{for} \ n \ \text{negative}. \\ \end{cases} $

Now if I expand this I get

\begin{align} f(t) \sum_{n=-\infty}^\infty c_n e^{int} & = \sum_{n=-\infty}^\infty c_n \bigg(\cos(nt) + i \sin(nt)\bigg) \\ & = c_0 + \sum_{n=1}^\infty c_n \bigg(\cos(nt) + i \sin(nt) + \cos(-nt) + i \sin(-nt)\bigg) \\ & = c_0 + \sum_{n=1}^\infty c_n \bigg(\cos(nt) + i \sin(nt) + \cos(nt) - i \sin(nt)\bigg) \\ & = c_0 + \sum_{n=1}^\infty c_n 2\cos(nt). \end{align}

But now the Fourier series has lost its dependence on sin functions! Am I incorrect or is this telling me that we only need $\cos$ functions to represent any periodic function using a Fourier series?

Answer 1

I rewrite your solution: \begin{align} f(t)& = \sum_{n=-\infty}^\infty c_n e^{int}\\ & = \sum_{n=-\infty}^\infty c_n \bigg(\cos(nt) + i \sin(nt)\bigg) \\ & = c_0 + \sum_{n=1}^\infty c_n \bigg(\cos(nt) + i \sin(nt) \bigg)+\sum_{n=1}^\infty c_{-n} \bigg(\cos(-nt) + i \sin(-nt)\bigg) \\ & = c_0 + \sum_{n=1}^\infty \color{blue}{c_n \bigg(\cos(nt) + i \sin(nt) + \cos(nt) - i \sin(nt)\bigg)} \\ & = c_0 + \sum_{n=1}^\infty c_n 2\cos(nt). \end{align} the blue line is true if $c_n=c_{-n}$ or by definition $(a - ib_n)/2=(a_{-n} + ib_{-n})/2$ or $b_n=0$ that means $c_n$'s are real, with other notation $$f(t)=\sum_{n=-\infty}^\infty c_n e^{int}=d+\sum_{n>0}^\infty c_n e^{int}+\sum_{n<0}^\infty c_n e^{int}=d+\sum_{n>0}^\infty( c_n e^{int}+ c_{-n} )e^{-int}=d+\sum_{n>0}^\infty (a_n e^{int}+ a_n e^{-int})=d+\sum_{n>0}^\infty a_n 2\cos nt$$

Answer 2

The cosines are even functions and any (possibly infinite) linear combination of even functions is itself an even function.

This means that if your function is not even, then cosines will not suffice. Conversely, it can be shown that if a function is even, then its Fourier series will contain no sine terms.