Because of the importance of sinusoidal functions, I will devote this section to a review of some of their properties.

Let us being with the sine wave,

$\begin{matrix} v(t)={{V}_{m}}\sin (\omega t) & \cdots & (1) \\\end{matrix}$

Which is sketched in figure (1). The amplitude of the sinusoid is V_{m}, which is the maximum value that the function attains. The radian frequency, or angular frequency, is ω, measured in radian per second (rad/s).

Fig.1: Sinusoidal Function

The sinusoid is a periodic function, defined generally by the property

$\begin{matrix} v(t+T)=v(t) & \cdots & (2) \\\end{matrix}$

Where T is the period. That is, the function goes through a complete cycle, or period, which is then repeated, every T seconds. In the case of the sinusoid, the period is

$\begin{matrix} T=\frac{2\pi }{\omega } & \cdots & (3) \\\end{matrix}$

As may be seen from (1) and (2). Thus in 1 second the function goes through 1/T cycles, or periods. Its frequency f is then

$\begin{matrix} f=\frac{1}{T}=\frac{\omega }{2\pi } & \cdots & (4) \\\end{matrix}$

Cycles per second, or hertz (abbreviated Hz). The relation between frequency and radian frequency is seen by (4) to be

$\begin{matrix} \omega =2\pi f & \cdots & (5) \\\end{matrix}$

A more general sinusoidal expression is given by

\[\begin{matrix} v(t)={{V}_{m}}\sin (\omega t+\phi ) & \cdots & (6) \\\end{matrix}\]

Where *ϕ *is the phase angle, or simply the phase. To be consistent, sinωt is in radians, *ϕ* should be expressed in radians. However, it is more convenient to specify *ϕ *in degrees. For example, we may write

$v(t)={{V}_{m}}\sin (2t+\frac{\pi }{4})$

Or

$v(t)={{V}_{m}}\sin (2t+{{45}^{o}})$

Interchangeably, even though the latter expression contains a mathematical inconsistencies.

A sketch of (6) is shown in figure (2) by the solid line, along with a sketch of (1), shown dashed. The solid curve is simply the dashed curve displaced ϕ/ω seconds, or ϕ radians to the left. Therefore, points on the solid curve, such as its peaks, occur ϕ radians, or ϕ/ω seconds earlier than corresponding points on the dashed curve.

Fig.2: Two Sinusoids with different phases

Accordingly, we shall say that \[v(t)={{V}_{m}}\sin (\omega t+\phi )\] leads \[v(t)={{V}_{m}}\sin (\omega t)\] by ϕ radians (or degrees). In general, the sinusoid

\[{{v}_{1}}={{V}_{m1}}\sin (\omega t+\alpha )\]

Leads the sinusoid

\[{{v}_{2}}={{V}_{m2}}\sin (\omega t+\beta )\]

By α-β. An equivalent expression is that v_{2} lags v_{1} by α-β.

**Example**

As an example, consider

\[{{v}_{1}}=4\sin (2t+{{30}^{o}})\]

And

\[{{v}_{2}}=6\sin (2t-{{12}^{o}})\]

Then v_{1} leads v_{2} (or v_{2} lags v_{1}) by 30-(-12) =42^{o}.

Thus far we have considered sine functions rather than cosine functions in defining sinusoids. It does not matter which form we use since

$\begin{matrix} \cos (\omega t-\frac{\pi }{2})=\sin (\omega t) & \cdots & (7) \\\end{matrix}$

Or

$\begin{matrix} sin(\omega t+\frac{\pi }{2})=\cos (\omega t) & \cdots & (8) \\\end{matrix}$

The only difference between sines and cosines is thus the phase angles. For example, we may write (6) as

\[v(t)={{V}_{m}}cos(\omega t+\phi -\frac{\pi }{2})\]

To determine how much one sinusoid leads or lags another of the same frequency, we must first express both as since waves or cosine waves with positive amplitudes,

For example, let

\[{{v}_{1}}=4\cos (2t+{{30}^{o}})\]

And

\[{{v}_{2}}=-2\sin (2t+{{18}^{o}})\]

Then, since

\[-\sin (\omega t)=\sin (\omega t+{{180}^{o}})\]

We have

$\begin{align} & {{v}_{2}}=2\sin (2t+{{18}^{o}}+{{180}^{o}}) \\ & =2\cos (2t+{{18}^{o}}+{{180}^{o}}-{{90}^{o}}) \\ & =2\cos (2t+{{18}^{o}}+{{108}^{o}}) \\\end{align}$

Comparing this last expression with v_{1}, we see that v_{1} leads v_{2} by 30^{o}-180^{o}=-78^{o}, which is the same as saying v_{1} lags v_{2} by 78^{o}.

The sum of a sine wave and a cosine wave of the same frequency is another sinusoid of that frequency. To show this, consider

$A\cos (\omega t)+B\sin (\omega t)=\sqrt{{{A}^{2}}+{{B}^{2}}}\left[ \frac{A}{\sqrt{{{A}^{2}}+{{B}^{2}}}}\cos (\omega t)+\frac{B}{\sqrt{{{A}^{2}}+{{B}^{2}}}}\sin (\omega t) \right]$

Which by figure (3), may be written as

$A\cos (\omega t)+B\sin (\omega t)=\sqrt{{{A}^{2}}+{{B}^{2}}}\left[ \cos (\omega t)\cos (\theta )+\sin (\omega t)sin(\theta ) \right]$

By a formula from trigonometry, this is

$\begin{matrix} A\cos (\omega t)+B\sin (\omega t)=\sqrt{{{A}^{2}}+{{B}^{2}}}\cos (\omega t-\theta ) & \cdots & (9) \\\end{matrix}$

Where, by figure (3),

$\begin{matrix} \theta ={{\tan }^{-1}}\frac{B}{A} & \cdots & (10) \\\end{matrix}$

Fig.3: Triangle useful in adding two sinusoids

A similar result may be established if the sine and cosine terms have phase angles other than zero, indicating that, in general, the sum of two sinusoids of a given frequency is another sinusoid of the same frequency.

https://www.youtube.com/watch?v=OGI-065RhFo

- You May Also Read: Phasors in Electric Circuits