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

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?

• Is $c_n=c_{-n}$.? Commented Mar 8, 2017 at 7:55
• @MyGlasses I have edited the post to specify the $c_n$'s. Commented Mar 8, 2017 at 8:02

• In the other answer to this question MyGlasses has derived another representation in which, yet again, $f(t)$ is only composed of combinations of $\cos$ functions. But that conflicts with your answer? Commented Mar 11, 2017 at 11:53
• The representation he derived is only true if $c_n=c_{-n}$. This happens only if $b_n=0$ for all $n>0$, which means the Fourier series contains no sine terms. Commented Mar 11, 2017 at 18:26
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 $$f(t)=\sum_{n=-\infty}^\infty c_n e^{int}=d+\sum_{n>0}^\infty c_n e^{int}+\sum_{n<0}^\infty c_n e^{int}=d+\sum_{n>0}^\infty( c_n e^{int}+ c_{-n} )e^{-int}=d+\sum_{n>0}^\infty (a_n e^{int}+ a_n e^{-int})=d+\sum_{n>0}^\infty a_n 2\cos nt$$
• By writing it in the other notation you have ended up with only cos functions also. So does that mean you have shown we only need cos functions to represent any periodic functions $f(t)$? Commented Mar 11, 2017 at 11:51
• This result obtain by $b_n=0$ and this will be only for even functions!. Commented Mar 11, 2017 at 11:56