Trig Function Identities with a Little Sound

By Brian Pasero

Reciprocal Identities : Sec, Csc and Cot

Let's start with Csc.  It is the reciprocal identity of Sin, that is to say:  Sinθ = (1/Csc)θand Cscθ = (1/Sin)θ.  
Here is a table of Sin θ values as θ runs from 0 to 2π.

nsinn=Table[{x, Sin[x]}, {x, 0, 2  π, π/12}]

And here's the plot of those points.

ListPlot[nsinn];

Here's the same table as above for Csc θ

ncscn=Table[{x, 1/Sin[x]}, {x, 0, 2  π, π/12}]

ListPlot[ncscn];

Look at the table below which compares the two functions.

({{θ, 0, (π/6), (π/4), (π/3), (π/2), ((2  π)/3), ((3  π)/4), ((5  π)/6), π, ((7  π)/6), ((5  π)/4), ((4  π)/3), ((3  π)/2), ((5  π)/3), ((7  π)/4), ((11  π)/6), 2  π}, {Sinθ, 0, (1/2), (1/(Sqrt[2])), ((Sqrt[3])/2), 1, ((Sqrt[3])/2), (1/(Sqrt[2])), (1/2), 0, -(1/2), -(1/(Sqrt[2])), -((Sqrt[3])/2), -1, -((Sqrt[3])/2), -(1/(Sqrt[2])), -(1/2), 0}, {Cscθ, ∞, 2, (Sqrt[2]), (2/(Sqrt[3])), 1, (2/(Sqrt[3])), (Sqrt[2]), 2, ∞, -2, -(Sqrt[2]), -(2/(Sqrt[3])), -1, -(2/(Sqrt[3])), -(Sqrt[2]), -2, ∞}})

How do the values compare to each other?

Here is a graph of Cscθ  as θ goes from 0 to 2π.

Clear[f, g, s, x, cscplot, secplot, cotplot]
f[x_] = 1/Sin[x];
cscplot  =  
        Plot[f[x], {x, 0, 2  π}, 
          PlotRange -> {{0, 2  π}, {-4, 
                4}}];

What is happening that causes the graph of csc to discontinue and picks up again below the x-axis?

This is the table for Cosθ

ncosn=Table[{x, Cos[x]}, {x, 0, 2  π, π/12}]

This is the Cosine plot.

Clear[f, g, s, x, secplot, cotplot]
f[x_] = Cos[x];
cosplot  =  
        Plot[f[x], {x, 0, 2  π}, 
          PlotRange -> {{0, 2  π}, {-4, 4}}];

Sec is the reciprocal identity to Cosine.  Use the table and plot to make a prediction about how the graph of secθ will look as θ runs from 0 to 2π.

Plot the graph of sec

Were you correct?  If not, why?

This is the table for Tanθ

ntann=Table[{x, Tan[x]}, {x, 0, 2  π, π/12}]

Here is the Tangent plot.

Clear[f, g, s, x, tanplot, cotplot]
f[x_] = Tan[x];
tanplot  =  
        Plot[f[x], {x, 0, 2  π}, 
          PlotRange -> {{0, 2  π}, {-4, 4}}];

Tan is the reciprocal identity to Cot.  Use the table and plot to make a prediction about how the graph of cotθ will look as θ runs from 0 to 2π.

Plot the graph of Cot.

Were you correct?  If not, why?

Pythagorean Identities:

Look at some of the values of (Sin^2) θ and (Cos^2) θ as θ runs from 0 to 2π.

{N[Sin[(π/6)]]^2, 
N[Cos[(π/6)]]^2}

{N[Sin[(π/4)]]^2, 
N[Cos[(π/4)]]^2}

{N[Sin[(π/3)]]^2, 
N[Cos[(π/3)]]^2}

{N[Sin[(π/12)]]^2, 
N[Cos[(π/12)]]^2}

{N[Sin[((7  π)/6)]]^2, 
N[Cos[((7  π)/6)]]^2}

What is the sum of each pair at the given θ's?

Would this be the sum for all θ's? (Try some values of your own)

If so, what Identity could you write to represent these sums for any value of θ?

Look at some of the values of (Tan^2) θ and (Sec^2) θ as θ runs from 0 to 2π.

{N[Tan[(π/6)]]^2, 
N[Sec[(π/6)]]^2}

{N[Tan[(π/4)]]^2, 
N[Sec[(π/4)]]^2}

{N[Tan[(π/3)]]^2, 
N[Sec[(π/3)]]^2}

{N[Tan[(π/12)]]^2, 
N[Sec[(π/12)]]^2}

{N[Tan[((7  π)/6)]]^2, 
N[Sec[((7  π)/6)]]^2}

How are the components of each pair related?

Would this relationship hold true for all values of θ? (Try some values of your own)

If so, what Identity could you write to represent this relationship for any value of θ?

Look at some of the values of (Cot^2) θ and (Csc^2) θ as θ runs from 0 to 2π.

{N[Cot[(π/6)]]^2, 
N[Csc[(π/6)]]^2}

{N[Cot[(π/4)]]^2, 
N[Csc[(π/4)]]^2}

{N[Cot[(π/3)]]^2, 
N[Csc[(π/3)]]^2}

{N[Cot[(π/12)]]^2, 
N[Csc[(π/12)]]^2}

{N[Cot[((7  π)/6)]]^2, 
N[Csc[((7  π)/6)]]^2}

How are the components of each pair related?

Would this relationship hold true for all values of θ? (Try some values of your own)

If so, what Identity could you write to represent this relationship for any value of θ?

Sound

Pitch (Frequency)

Where do the graphs of sinusoids show up?  Sound is one place.  Sound is a wave that can be can be created in air, water and other substances.  The wave is an indicator of the pressure that you ear picks up.  You begin to hear sound when the pressure changes 20 times in a second (although this is very tough to hear).  Here is an example of one cycle of pressure change in one second.

Plot[Sin[2 π (t)], {t, 0, 1}, 
        PlotStyle -> {{RGBColor[1, 0, 0], Thickness[0.01]}}, 
        AxesLabel  ->  {"Time", "Pressure"}, 
        AspectRatio ->  1/GoldenRatio] ;

Here the pressure went from 0 to 1 to -1 and then back to 0 again.  Here is a graph of two complete cycles occuring in two seconds.

Plot[Sin[2 π (t)], {t, 0, 2}, 
        PlotStyle -> {{RGBColor[1, 0, 0], Thickness[0.01]}}, 
        AxesLabel  ->  {"Time", "Pressure"}, 
        AspectRatio ->  1/GoldenRatio] ;

>From this you can see that the pressure will continue to occsilate between 1 and -1 for as long as we allow.  However humans cannot hear only one cycle per second.  As mentioned before humans need at least 20 cycles of pressure change per second before they begin to here anything.  Here is what that sounds and looks like.

Clear[f, t];
f[t_] = 20 ;
FMPlay[f[t], {t, 0, 1}];

The sound about has a frequency of 20 Hz because that is the number of cycles of pressure change that occurs in one second.  Frequency is often referred to as Pitch and is measured in Hz.  Human range of hearing is between  20 and 20,000Hz.  Dogs can hear a higher frequeny than humans.  
Here is graph and sound of 5,000 Hz.

Clear[f, t];
f[t_] = 5000 ;
FMPlay[f[t], {t, 0, 1}];

That didn't sound so good.  Lets tone it down a little to a more reasonable sound.

Clear[f, t];
f[t_] = 1000 ;
FMPlay[f[t], {t, 0, 1}];

Solid!  In the last two examples the graphs didn't turn out as well as they did when the frequency was at 1 and 20.  
What could be the cause of this?

Also, since frequency is related to the number of cycles per second of pressure difference, (shown more clearly in the first two graphs) how is the pitch related to the general form of the equation of the sinusoid.  f[x] = D + Asin(Bx - C) ?

Make an adjustment to the following code that would start low and end on a high note.  (Try not to get too carried away though, those high frequencies can put a strain on the ears.)

Clear[f, t];
f[t_] = 20 (t^2);
FMPlay[f[t], {t, 0, 5}];

Amplitude

The amplitude of sound is its volume.  The higher the amplitude the louder the sound.  Lets look at some sounds with different amplitudes.  The first sound has an amp. of 3, the second sound has an amp. of 1.  Take a listen.

Clear[sound1, sound2, sound3, sound4];
sound1 = 
        AMPlay[3, {t, 0, 3}, PlayRange ->  {-3, 3}];

sound2 = 
        AMPlay[1, {t, 0, 3}, PlayRange ->  {-3, 3}];

Here is a function that allows the amplitude to start small and grow.

Clear[f, t];
f[t_] = (1/((t - 10)));
AMPlay[f[t], {t, 5, 9.5}];

Look closely at the function f[t] that delegates the above amplitude.  Why does the volume increase as t increases?



Finally, in the sound below we have both the pitch and the amplitude changing.  By simply looking at the function f[t], playing around with the exponent of (t + 1) and using your existing knowledge about the graphs of sinusoids, discuss what parts of the function makes the volume change and what makes the frequency change.

Play[
        (E^(-t)) Sin[2 π 220 (((t + 1))^((7/2)))], {t, -1, 
          4}];

This might help the brainstorming process.

Clear[f, x]
f[x_] = Sin[(x^2)];
Plot[f[x], {x, 0, 2  π}];

The information on sound was obtained from the University of Illinois Math Department's sound.nb file.  Check it out to learn more about sounds and sound waves.