# Why a sinusodial signal expressed by a cosine function instead of a sin function?

In the signals and systems course one of the first things that we see is a sinusodial signal wave. And they say its expressed like $A\cdot\cos(\omega t + \phi)$. But why we use a cosine function when we work on a sinusodial signal? why its not $A\cdot\sin(\omega t + \phi)$?

or watch the first minute of this video: https://ocw.mit.edu/resources/res-6-007-signals-and-systems-spring-2011/video-lectures/lecture-2-signals-and-systems-part-i/

• Mathematically, you can use either sin or cosine. The shape of the waveform is the same in either case. It is just shifted by 90 degrees. Commented Mar 26, 2017 at 21:28
• There are cases where Cos is preferred over Sin. Displacement powerfactor, magnetic alignment ... Really depends on whether 0deg has a special meaning.
– user16222
Commented Mar 26, 2017 at 21:38
• @JonRB, not sure if you are addressing me. But when I said you can use either one, I just meant that they are both correct. I think cosine is just more convenient. Commented Mar 26, 2017 at 21:47

A cosine and sine wave are essentially phase-shifted versions of one another. For instance,

\begin{equation*} cos(x+\phi) = sin(x+\phi-90) = sin(x+\theta) \end{equation*}

where

\begin{equation*} \theta = \phi-90 \end{equation*}

Hence, a cosine and sine wave are essentially the same, and what differentiates their use is the value of the initial phase.

However, a primary reason as to why the cosine notation is preferred is because of the frequent occurrence of complex envelopes in the area of Signals/Systems. An example of the use of complex envelopes is when a sinusoid of low frequency, say m(t), is modulated onto a sinusoid of higher frequency (fc); the resultant modulated signal r(t) can be expressed as:

\begin{equation*} r(t) = Re\{m(t)e^{j 2 \pi f_{c} t} \} \end{equation*}

As it turns out, the real part of a complex envelope is a cosine waveform (Refer to Euler's formula) and hence it is much more convenient to represent a signal, such as r(t), using a cosine.

Yes, the most common reason given is that a cosine is the real part of a complex exponential in Euler's formula, and thus implies Euler's dentity.

But why is a cosine the real part?

One good reason is that a cosine allows a signal with a phase and frequency of zero to have a non-zero and real DC value. There are several other interesting aspects. One is that the cosine is symmetric around the origin (0), and has a 1st derivative of zero there. In practical electronics, that means that time is reversible around the origin, and that a small phase error from a starting phase of zero will have the minimal effect on some circuit behaviors.

• The Fourier transform also spits out phases relative to the cosine, not the sine. I don't know why, but it does. If you construct an artificial waveform by adding up a bunch of different frequency sinewave where one of them has a phase of zero, the phase spectra produced by the transform will assign a 90 degree phase to that zero degree frequency (aka a zero phase cosine), and all the other sine waves will have a 90 degree phase shift as well. Commented Aug 9, 2021 at 5:03

The general expression for an oscillating system relies on the relationship $$e^{j{\omega}t} = cos(\omega t) + j\text{ } sin(\omega t)$$ The cosine term is the real part, while the sine term is the imaginary. Physical measurements can only directly measure real quantities (barring special devices), so such systems are generally characterized by variables which are expressed as cosines.