Chapter 7, Sections 3 and 4



Math 1330, Chapter 5, Sections 1 and start of 2

A more modern approach to trigonometry is to use real numbers rather than angles as the inputs (or domain). We find those real numbers by noting that the arc length subtended by a rotation from standard position on the unit circle is the same number as (, the rotation itself. Conventionally, we switch from ( to “t” when we do this

If we start at standard position and t = 0, we can create a number line wrapping in the positive direction and the negative direction…

We can find the whole number line wrapped around the unit circle. As I told you before, we will then associate the functions (cos t, sin t) with P a point on the unit circle. This convention gives you the function values for both sine and cosine simultaneously.

Note that sin x ( x in general, nor is cos x the same thing as x.

Opposite Angles:

Let’s take a minute to explore the unit circle idea further….for each input value t, there’s another input (t…it turns out that the sine and cosine of (t is related to those function of t. Here’s how

[pic]

Let’s look at some of the quadrant one reference angles and their quadrant 4 counterparts.

cos [pic] cos ( [pic]

sin[pic] sin ([pic]

indeed let’s look at the signs on the values for P a point in Q1 vs a P located symmetrically in QIV…

so we find that cosine is an even function [pic]

and sine is an odd function (along with tangent) [pic]

These are terrifically handy identities and will be used fairly constantly throughout the rest of your math career; please know the vocabulary and the effect by heart.

Simplify:

[pic]

[pic]

[pic]

If sin t = .75, what is sin ((t)?

If cos t = 1/5, what is cos ((t)?

Periodicity:

Both the sine and cosine functions are periodic with period 2π. Thus if you use an even number of whole rotations you’ll arrive at the same point P on the circle and your trig function values will be the same

sin ( t + 2nπ) = sin t where n is any integer

cos ( t + 2nπ) = cos t

Let’s check this out. Let n be (5. If we’re at a point P on the unit circle and we rotate 2 ((5) π times around the circle what are the values for sine and cosine:

[pic]

Simplify

[pic]

[pic]

If cos t = (.33, what is cos (2π(t)?

If sin t = .11, what is sin (t + 200π)?

cos ([pic]

sin ( [pic]

Now let’s work some problems

If sin t = [pic] and [pic], find cot t.

If cot ( = ( [pic] and cos ( < 0 what is csc ( ?

If cos x = (1/3 and sin x > 0, what is tan x?

Graphs of the sine function and the cosine function:

In general, periodic functions have a regular, dependable, repeating shape. There are several nice examples periodic functions in your text. All that is required with a periodic function is that you know the graph well on one period…then you just copy it over and over to the left and to the right.

For the trigonometric functions, we’re going to look at the

• period,

• the amplitude,

• the phase shift and

• vertical shifting.

Amplitude is defined by the range values. Amplitude is defined to be half of the absolute value of the difference between the maximum and the minimum in one period.

Here’s an example:

For sines and cosines:

Let’s mark off an x axis with the quadrantal angles in radian measure including some negative rotations. We know that both sine and cosine are periodic with period 2π.

Note where the quadrants are and note that going left from zero puts us into Quadrant 4 again. The quadrantal angles mark off the 4 parts of the period for both trig functions.

Now let’s put in the y values for these quadrant markers.

| |0 |[pic] |[pic] |[pic] |([pic] |

|sin t | | | | | |

|cos t | | | | | |

| | | | | | |

Let’s graph the sine function from ([pic] to [pic] .

Put some marks where the reference angles are.

Note then that the sine function, f(t) = sin t is

periodic with period 2π,

has an amplitude of one and

starts the period at x = 0;

the domain is all Real numbers and

the range is [(1,1].

What’s the y intercept ? How many of these? How come?

How about the x intercepts?

We have 5 parts to the standard period:

1. (start, mid)

2. (Q1, max)

3. (mid, mid)

4. (Q3, min)

5. (end, mid)

One more look:

Now let’s do the same for y = cos t

First let’s graph the cosine function from ([pic] to [pic]. Put in marks for the reference angles.

So, the cosine function is

periodic with period period 2π,

has an amplitude of one and

starts the period at x = 0;

the domain is all Real numbers and

the range is [(1,1].

What’s the y intercept?

Some x intercepts?

There are 5 parts to the standard cosine period

1. (start, max)

2. (Q1, mid)

3. (mid, min)

4. (Q3, mid)

5. (end, max)

One more look:

Transformations with sine and cosine graphs:

The expanded sine and cosine functions:

y = A sin(Bt + C) + D y = A cos(Bt + C) + D

On the basic or standard sine and cosine functions we have

A = 1

B = 1

C = 0

D = 0

f(x) = sin x f(x) = cos x

A is the amplitude. It tells you the value for the max and min of the function

B is the period adjustor.

The period of sine or cosine is

[pic]

C and B together as a ratio spell out the phase shift – the basic sine and cosine start the standard period at x = 0. But there are shifted sines and cosines that start at other places. Note that C is zero with the standard functions.

D indicates the midvalue of the range. The basic graphs have a mid range value of y = 0, the x axis. But it’s not always so “+ 1” added on the end…shifts the graph up 1 just as you’d expect. Note that D is zero with the standard functions.

Let’s look at what happens when we start working with A’s and B’s that are not 1.

In the next 3 examples, C and D are both zero.

What’s the function and what’s the amplitude?

What’s the equation?

What’s the function and what’s the period?

hint, the point ([pic]is on the graph.

What’s the equation?

Ok, now for a tough one:

What’s the equation?

What about when C and D are NOT zero?

Here’s one of those:

f(x) = [pic]

Continuted:

the graph of f(t) = A sin(Bt + C) and A cos(Bt + C)

Amplitude changes (B = 1, C = 0, D = 0)

A = 1 standard graph

(1A reflect x axis

[pic]= k max is at k, min is at (k

there are no changes to the starting point nor the quarter points of 1 period

the range changes to [(k, k] from [(1, 1]

graph one period of y = 2sin t

amplitude is range is

no reflection

B =

C =

start

Q1

mid

Q3

end

graph one period of y = [pic]

amplitude is range is

reflect about the x axis !

B =

C =

start

Q1

mid

Q3

end

Period changes (A = 1, C = 0, D = 0)

standard period: 2π

new period length [pic]

new quarter points! take the original period 0< t < 2π and divide through by B.

Then calculate the new quarter period points and graph.

Note that for [a,b] the midpoint is [pic]

graph y = cos 2t unreflected cosine shape

Amplitude is range is

new period length is

C =

start

Q1

mid

Q3

end

graph y = sin 2π t unreflected sine shape

Amplitude is

new period length is

C =

start

Q1

mid

Q3

end

Phase shift (A = 1, B = 1, D = 0)

C changes the start and stop points for a standard period. C doesn’t work independently from B, though.

The new starting point is [pic] [pic] shifting left or right depending on the sign of C

and the new ending point is [pic][pic]

graph y = sin (t + [pic])

this is a sine shape that is not reflected

amplitude is range is

the maximum is and occurs where in the period

the minimum is and occurs where in the period

new period is

phase shift is R or L? how far?

Let’s do some work calculating the important domain points for one period

reminder:

Note that for [a,b] the midpoint is [pic]

The period will start at The period will end at

The mid point of the period is

Q1 is

Q3 is

the y intercept is

List these 6 important points. Graph.

Working these three graph shifting variables together

Graph y = sin (2θ ( [pic])

reflection about x axis? yes or no

A = range is

the max is and occurs where in the period

the min is and occurs where in the period

B =

so new period is

starts at ends at

midpoint

Q1

Q3

the y intercept is

And another one f(t) = ( cos (πt +[pic])

reflection!

A = range is

max is and occurs where in the period

min is and occurs where in the period

B =

new period length

shifts left or right by how much

start point is end point is

midpoint

Q1 Q3

y intercept:

Graph the function

And a tougher one f(t) = 5 sin ( [pic]t ( [pic])

Amplitude range is

max is where in the period does it happen

min is where in the period does it happen

new period length

shift left or right?

starting point ending point

midpoint Q1 Q3

y intercept

graph

graph y = (0.5 cos (3t)

reflection?

amplitude is range is

max is occurring where

min is occurring where

new period length is

starting where ending where

midpoint Q1 Q3

y intercept

graph two periods. Show 9 important points.

True or false. Tell why or why not, too.

cos (3t) = 3cos t

sin ((t (2π) ’ ( sin t.

The domain of y = sin t is [0, 2π).

The y intercept for y = 3 sin (2t ([pic] is ([pic].

(1, [pic]) is a graph point of y = (2cos(πt + [pic]).

y = sin t and y = cos t do not intersect at all.

sin t and csc t do not intersect.

cos t and sec t intersect at infinitely many points.

Graphing outline

A sin/cos (Bt + C)

function

A = range is [ ]

reflection: no yes (circle one)

B = npl =

phase shift: none left right (circle one)

start point

end point

mid

Q1

Q3

y intercept

Graph

The graphs of tangent and the reciprocal functions: cosecant and secant:

Here’s the easiest way to graph the tangent function:

Translate the graph to it’s rational format:

Lightly sketch in the cosine function:

everywhere cosine has an x intercept put in a vertical asymptote….why?

put in points where the sine function has t axis intercepts (integer multiples of [pic]) these will be the intercepts for tangent, too. Why?

Recall that in Q1 and Q3 tangent is > 0 and in Q2 and Q4 tangent is < 0.

Indeed in Q1 and Q3 what’s the value of the tangent of angles related to [pic]?

And in Q2 and Q4? perfect…put those points in the appropriate places. Connect them smoothly and Voila!

let’s run back through this:

we got the vertical asymptotes from?

we got the t axis intercepts from?

we got nice graphable points from?

Now tangent and cotangent look alarmingly alike…and you’re expected to know these graphs by heart.

Let’s follow a similar plan to sketch in cotangent:

Translate the graph to it’s rational format.

Which function do you sketch on the t axis to find the vertical asymptotes?

Which function are we going to use to get the t axis intercepts? where are they and why are they there?

What are the very nicest values for cotangent t and where are they? Put them on the graph and smoothly connect. Done.

Note that the period for tangent is length π unlike sine or cosine. Thus if you encounter a B that is not 1, you find the new period length by using [pic].

Now what about the other reciprocal functions?

Cosecant is the reciprocal for sine.

Secant is the reciprocal for cosine.

We’re going to build them from sines and cosines with the Snip and Flip method.

For y = csc t

first sketch in sin t

Note two facts:

everywhere sin t has an intercept csc t has a VA

at the max and min points of the period sin t = csc t

Snip the sine function at every intercept and flip it about the shared point with csc t.

What about graphing csc 2t? Graph sin 2t, snip and flip.

What about graphing 2 csc t? Graph 2sin t, snip and flip.

What’s the difference in handling csc t and sec t?

How about sec t?

What’s your best guess for the easy way to do it?

What about y = sec πt?

What about y = π sec t?

List 3 vertical asymptotes for [pic]

List 3 vertical asymptotes for [pic]

Tell the amplitude, period, the y intercept, and 3 x intercepts for [pic].

Graph [pic]

Ok, now for some pattern recognition work. I’ve shown one period for each function. Answer the following questions

What’s the equation that gives us:

What’s the equation?

What are the asymptotes and min/max if this is the graph of y = 3 csc (2t)?

What’s the equation?

([pic]

What’s the equation?

Inverse functions:

Recall 1 to 1 functions have inverse functions

if ( a, b ) is on a 1 to 1 function’s graph, then ( b, a ) is on it’s inverse function’s graph

[pic]

So let’s start with the sine function and restrict it’s domain.

[Let’s review that we’re graphing [pic]and not csc x!!!]

the restricted domain is

the resultant range is

Let’s make a table of points for Q4 (expressed with negative rotations) and Q1

The domain of arcsin is

The range of arcsin is

The graph looks like:

For arccos(x) = [pic] [again NOT secant x]

Let’s look at what we do to the cosine function so it’s 1 to 1

The restricted domain is

The range is

Let’s make a table of graph points for arccos(x)

Here’s the graph:

the domain is

the range is

Now for tangent:

The original graph looks like:

to get it to be a 1 to 1 function we will

So for arctan x = [pic], the domain is

the range is

the graph looks like:

Know these by heart, please.

Now let’s start solving some problems. First graph points:

[pic] hint: is this in the domain?

[pic]

[pic]

sin ([pic]

tan ( [pic]

cos ([pic]

cos ( [pic]

sin ([pic]

tan ( [pic]

New identity equation

[pic]

Let’s look at why this is true

Now let’s do some quick reviewing:

Sketch the graph for [pic]

Sketch the graph for [pic]

Sketch the graph for [pic]

Now sketch the graph of y = [pic]

Now sketch the graph of y = [pic]

Supply the following values

sec([pic]

tan ([pic] hint: [pic]

sin ( cos[pic]

sin([pic]

cos([pic]

tan ( [pic]

-----------------------

0

pð

(1.5, (2)

[pic]

[pic]

[pic]

[pic]

(3pð,ð(5ð)ð

π

(1.5, (2)

[pic]

[pic]

[pic]

[pic]

(3π,(5)

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