## Monday, October 31, 2011

### Phase shift

We have dealt with three ways to change a trig function.

1) Amplitude
f(x) = asinx or g(x) = acosx

This changes the height of the graph.  Instead of oscillating between -1 and 1, the high and low points will be |a| and -|a|.  (It could be that a is negative, so we need the absolute value signs.)

2) Frequency (or period)

f(x) = sinpx or g(x) = cospx

This changes the distance between the high points.  For sine and cosine, the period is 2pi.  This changes to |2pi/p|.  If |p| > 1, it takes a shorter distance to repeat.  If |p| < 1, the wave is stretched and takes a longer distance to repeat.

3) Adding a constant
f(x) = c + sinx or g(x) = c + cosx

This makes a graph fluctuates between c - 1 and c + 1.  If there is an amplitude multiplier a, the fluctuation is between c - |a| and c + |a|.

The last way to change a trig function is called phase shift.

f(x) = sin(px+h) or g(x) = cos(px+h)

The "important" points in a trig function happen at multiples of ½pi.

sinx = 0 when x = ...-2pi, -pi, 0, pi, 2pi...
sinx = 1 when x = ... -3½pi, -1½pi, ½pi, 2½pi, 4½pi....
sinx = -1 when x = ... -2½pi, -½pi, pi, 3½pi, 5½pi....

cosx = 0 when x = ...-1½pi, -½pi, ½pi, 1½pi...
cosx = 1 when x = ... -2pi, 0, 2pi, 4pi....

cosx = -1 when x = ... -3pi, -pi, pi, 3pi, 5pi....

If we want to look at multiples of  ½pi, we let k be any integer and consider kpi/2.

px+h = kpi/2
px = kpi/2 - h
x = (kpi/2 - h)/p

These will be the values of x where the graph will be either in the middle or at the maximum or minimum.  The easiest thing to do is plug in k = 0, k = 1 and k = 2 in a row to see what you get. That means

x = - h/p
x = (pi/2 - h)/p
x = (pi - h)/p

There are four possible situations you will get plugging these numbers in.

Middle, high, middle
High, middle, low
Middle, low, middle
Low, middle, high

Once you have three such points, you can draw the entire graph.

For the following functions, find three points in a row where the function either reaches an extreme value or a median value.

f(x) = -3 + 2cos(pi(x + ¾))
g(x) = -4sin(2x - 5)
k(x) = cos(2 - 3x)

Answers in the notes.

## Monday, October 24, 2011

### The inverse trig functions

Here is the graph of tangent inverse, also called Arctan.

Domain: all real numbers (think of these numbers as the slope of a line.)
Co-domain: (-pi/2, pi/2) (These numbers refer to angles in the first and fourth quadrants, as well as the angle 0, which lies between the two quadrants.)

Domain: [-1, 1]
Co-domain: [-pi/2, pi/2]

Domain: [-1, 1]
Co-domain: [0, pi]

## Sunday, October 23, 2011

### Practice for graphing sine waves.

If you want to visualize the graphs you are asked to draw, the website Wolfram|Alpha has an instruction you can type called "plot". Here is the example of plot 3sin(2x/3).

Here are some examples for practice without graphs. You can do that yourself on Wolfram|Alpha. The answers are in the comments.

a] f(x) = 2 + 4sin(x/2)

period = ______

amplitude = _______

maximum of f is ______, reached when x = _______.

minimum of f is ______, reached when x = _______.

b] k(x) = 4 + 2cos(3x)

period = ______

amplitude = _______

maximum of k is ______, reached when x = _______.

minimum of k is ______, reached when x = _______.

c] q(x) = 1 - 3sin(5x)

period = ______

amplitude = _______

maximum of q is ______, reached when x = _______.

minimum of q is ______, reached when x = _______.

Answers in the comments.

## Monday, October 17, 2011

### practice with fractions of a circle

Consider 17/19 of the circle as an angle, which we will call theta.

a) Write theta as an angle in decimal degrees, rounded to four places.
b) Write theta in DMS, rounded to nearest degree.
c) Write theta as an exact fraction of the circle using pi.
d) Write theta as radians written without pi, rounded to four places after the decimal.
e) sin theta
f) cos theta
g) tan theta
h) cot theta
i) sec theta
j) csc theta

Answers in the comments.

## Sunday, October 16, 2011

### More practice solving trig equations.

Find the values of x between 0 and 2pi where these equations are true, written in radians as multiples of pi.

a) tanx - 6 = 0
b) sinx - ½tanx = 0
c) cos²x = ½

Answers in the comments.

## Friday, October 14, 2011

### The graphs of trig functions

Here are pictures of the graphs of all six trigonometric functions from 0 to 2pi, which is about 6.2832.  I borrowed this from a website and I have to say the picture of tangent does not look quite right. If you put a mirror on cotangent and shift it over, it should look exactly like tangent.  The little"bend" you see in the cotangent graph (known as an inflection point) is more accurate.

We will be discussing how to manipulate the functions in the next few classes.

## Tuesday, October 11, 2011

### "Famous" angles practice

The "famous" angles in the first quadrant (and on the x-axis and y-axis adjacent to the first quadrant are 0°, 30°, 45°, 60° and 90°. You are expected to know the sine, cosine and tangent values for all these angles.  When we move to the other quadrants, any angle that is 90° or 180° or 270° more than a famous angle will share trig values with one of the first quadrant angles or the value will be the negative.  The mnemonic device is
All (all trig function are positive in the first quadrant)
Students (only sine is positive in the second quadrant)
Take (only tangent is positive in the third quadrant)
Calculus (only cosine is positive in the fourth quadrant)

The three "minor" trig functions, secant, cosecant and cotangent can all be defined as reciprocals of major trig functions.

secx = 1/cosx
cscx = 1/sinx
cotx = 1/tanx

Here are six problems dealing with "famous" angles. Remember that pi radians = 180°.

a) cos(630°)
b) sin(-7pi/3)
c) tan(7pi/6)
d) cot(-420°)
e) sec(225°)
f) csc(13pi/3)

Answers in the comments.

## Friday, October 7, 2011

### solving for values of sine, cosine and tangent.

Let's assume we can solve a trigonometric equation to get a value for sina, cosa or tana.  We need only use the corresponding inverse function to get the angle a

For any of these simple type of problems, we are only half way home if we are required to find all the angles between 0° and 360°, or if we are measuring in radians, between 0 and 2pi.

Let's assume for argument's sake that the angle you get is a in our picture, something in first quadrant. (It's possible if cosa is negative, the first angle we will get is in the second quadrant.  If cosa < 0  or tana < 0, the angle will be negative degrees or radians and in the fourth quadrant.

The case for cosine:  If cosa = constant, then cos(360-a)° = constant. (If you are dealing with radians, it's a and 2pi - a.)

The case for sine: If sina = constant, then sin(180-a)° = constant. (If you are dealing with radians, it's a and pi - a.)

The case for tangent:  If tana = constant, then tan(180+a)° = constant. (If you are dealing with radians, it's a and pi + a.)

Problems. Give the answers as decimal degrees, rounded to the nearest thousandth of a degree and radians as k(pi), where k is rounded to four places after the decimal.

1) sina = .9
2) tanb = -2.5
3) cosc = sqrt(8)/3
*4) sinx = tanx

Answers in the comments.

## Monday, October 3, 2011

### Degrees (including DMS) and radians (with or without multiples of pi.

We are now going to refer to the measure of angles in two different ways.  When we discuss angles of a triangle, for example, it is standard to give them in degrees and use the rule that the sum of the interior angles is 180°. It's very important to always have a degree sign on a number referring to degrees, because 30° is certainly not 30 meters or 30 feet or 30 of any measure we use for distance.

When we think of the trigonometric functions cosine and sine, we get numbers between -1 and 1 that correspond to the x and y values of a point on the unit circle. Instead of measuring an angle in degrees, when the trig functions are used in calculus and other settings, the angle is defined by arc length instead of degree.  The idea is that the distance around the unit circle (circumference) is equal to 2pi, where pi is the number your calculator represents as 3.141592654...  Here are some well known angles written in degrees and as multiples of pi and rounded to four places after the decimal.

360° = 2pi ~= 6.2832

270° = 3pi/2 ~=4.7124

180° = pi ~= 3.1416
120° = 2pi/3 ~= 2.0944
90° = pi/2 ~= 1.5708
60° = pi/3 ~=1.0472
45° = pi/4 ~=0.7854
30° = pi/6 ~= 0.5236

The formula to change from a° to radians is to multiply by the fraction pi/180.

One way to think of this is if you were to walk about 6.2832 meters around a circle with a 1 meter radius, you would come back to your original starting place.  This means 1 radian is a little bit less than 1/6 the way around the circle.  Let's turn "nice" radian numbers into degrees to the nearest thousandth. (The formula is to multiply by the fraction 180/pi.)

1 radian ~= 57.2958°
2 radians ~= 114.5916°
3 radians ~= 171.8873°
4 radians ~= 229.1831°
5 radians ~= 286.4789°
6 radians ~= 343.7747°
7 radians ~= 401.0705° or 41.0705°

At 7 radians, we have traveled more than 360°, so we can subtract 360° to get an angle that is easier to read.

If I write a decimal degree rounded to four places after the decimal, this is to the nearest ten thousandth, which is very small slice of an entire circle.  Another way to write degrees that are not whole numbers is degrees-minutes-seconds or DMS.  A minute is 1/60 of a degree and a second is 1/60 or a minute. (1/60)(1/60) is 1/3600, so if we write an angle using this method, it is not quite as precise as rounding to four places after the decimal (nearest 1/10,000), but more precise than three places after the decimal (nearest 1/1,000).

In class, I showed a way to do these by hand, but I missed that the calculator can do this for us.  There is a button (third button, second row from the top) with the symbols ° ' ".  If I want to change 57.2958° to DMS, I can type in 57.2958, press the [° ' "] button and scroll all the way to the right to find |>DMS instruction.  When I press [ENTER] the answer line says

57° 17' 44.9"

Because this rounds to the nearest tenth of a second, this is slightly more precise than four places after the decimal, but not as precise as five.

Also, if an angle is given in DMS, we can change back to decimal degrees by using the symbols °, ' and ".  This way I can type in 57° 17' 44.9", press [ENTER] and get 57.2958° back.

If I type in the formula to change a single radian to a degree I type 180/pi = 47.29577951...; asking for DMS of this gives us

57° 17' 44.8"

So we can see there was some rounding error at four places after the decimal.  If this is typed in and [ENTER] is pressed, we get

57° 17' 44.8" ~= 57.29577777..., which is not exactly the number we typed in.

Problems

Write these fractions of the circle as

decimal degrees (rounded to four places after the decimal)
DMS (rounded to nearest second)
radians as a multiple of pi (rounded to four places after the decimal)
radians (rounded to four places after the decimal)

a) 11/16 of the circle
b) 13/25 of the circle
c) 7/50 of the circle

Answers in the comments.

### Two different versions of the Trigonometric Identity using the Pythagorean Theorem.

In my class, I call

sin²alpha + cos²alpha= 1

The Trigonometric Identity.

Other sources say there are many trigonometric identities, but the vast majority of them are just re-workings of this formula.  Here are two examples that look different but are very closely related.

Consider the following lengths of the sides for a right triangle

a = sinalpha, b = cosalpha, c = 1

These are the side lengths in what I call the Normal Standard Position Right Triangle, where the right angle is the lower right and alpha is the lower left and the hypotenuse has length 1.

What if we have a similar triangle, but instead of the hypotenuse having length 1, side b was length 1?  We can do this by dividing all the side lengths by cosalpha.

a = sinalpha/cosalpha, b = cosalpha/cosalpha, c = 1/cosalpha

or

a = tanalpha, b = 1, c = secalpha

Since this is still a right triangle, we get a new version of the Pythagorean Theorem

tan²alpha + 1 = sec²alpha

often re-written as

tan²alpha = sec²alpha - 1

The other possibility is to have a have length 1.  We do this by dividing all length by sinalpha.

a = sinalpha/sinalpha, b = cosalpha/sinalpha, c = 1/sinalpha

or

a = 1, b = cotalpha, c = cscalpha

This gives us yet another version of the Trigonometric Identity

cot²alpha + 1 = csc²alpha

often re-written as

cot²alpha = csc²alpha - 1

This brings us to the Complementary Identity rule.  If you have any trig formula, you can replace every trig function with its complement and you have another trig formula.  For example,

tanalpha = sinalpha/cosalpha

Change tan to cot, sin to cos and cos to sin and we get

cotalpha = cosalpha/sinalpha