Example similar to the take home final

By request, I present an example like the take home final.  This is the graph of r =2cos(2theta).  I calculated the values at all 24 points that have lines radiating from the origin and marked them, then I played connect the dots, smoothing the curve instead of drawing straight lines.  Here are the values at all 24 points.

r = 2cos(2*0) = 2

r = 2cos(2*pi/12) = sqrt(3)

r = 2cos(2*pi/6) = 1

r = 2cos(2*pi/4) = 0

r = 2cos(2*pi/3) =-1

r = 2cos(2*5pi/12) =-sqrt(3)

r = 2cos(2*pi/2) =-2

r = 2cos(2*7pi/12) =-sqrt(3)

r = 2cos(2*2pi/3) =-1

r = 2cos(2*3pi/4) = 0

r = 2cos(2*5pi/6) = 1

r = 2cos(2*11pi/12) = sqrt(3)

r = 2cos(pi) = 2

r = 2cos(2*13pi/12) = sqrt(3)

r = 2cos(2*7pi/6) = 1

r = 2cos(2*5pi/4) = 0

r = 2cos(2*4pi/3) =-1

r = 2cos(2*17pi/12) =-sqrt(3)

r = 2cos(2*3pi/2) =-2

r = 2cos(2*19pi/12) =-sqrt(3)

r = 2cos(2*5pi/3) =-1

r = 2cos(2*7pi/4) = 0

r = 2cos(2*11pi/6) = 1

r = 2cos(2*23pi/12) = sqrt(3)

The result is called a four leaf rose.With the problem from the take home, because we have the formula r = 2 + cos(3*theta), the radius will never get to be less than zero, so it will not cross the origin mulitple times like this one does.

The take home section of the final

Here is the take-home section of the final as a picture file.  You should be able to download it, but if not, send me an e-mail at mhubbard(at)peralta(dot)com, substituting the correct symbols for the words in parentheses.

I can also send you a list of the topics for the final if you did not get it in class today.  You are allowed to have notes on two 3" x 5" cards.

See you on Wednesday, Dec. 14 at 8:00 am, one hour earlier than usual.

Polar coordinates.

In class Monday, we introduced the idea of polar coordinates.  In rectilinear coordinates (x, y), the first number tells us how far right (positive) or left (negative) we move on the horizontal axis and the second number tells us how far up (positive) or down (negative) we move on the vertical axis.

In polar coordinates we have (r, theta).  There is a central point, not unlike (0, 0) in the xy-axis system, and we consider that we are pointing in a direction we call angle 0 in radians.  The standard is to make that direction to the right, the same as the positive x-axis in rectilinear.  r is the distance from the origin and theta is the angle given in radians.

It is acceptable to have negative values for distance and for angles.

Unlike rectilinear coordinates, polar coordinates are not unique.  The easiest example of this are the coordinate pairs (0, 0) and (0, 1).  The first zero means we are a distance of 0 from the origin.  That means we are on the origin.  The angle we turn doesn't change where we are.

Even when points aren't the origin, the polar coordinates are not unique. The simplest example is

(1, 0) = (1, 2pi)

What this says is if we are 1 away from the origin and pointing to the right, this is the same as being 1 away from the origin and turning a full circle. In fact, adding or subtracting any multiple of 2pi to an angle brings us back to pointing in the exact same direction.

We will be working more in polar coordinates for the rest of the week, possibly longer.

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step by step through some half angles of famous angles.

The half angle formula deal with fractions and square roots.  The Blogger text software doesn't have good ways to type those in, so I resort to using pictures with captions.

These are the original half angle formulas.

This is a step by step (across) way to turn the sine and cosine of 15 degrees into fractions with whole number denominators.  These answers look different from the ones you get using subtraction of angles, either 60 degrees - 45 degrees or 45 degrees - 30 degrees.  If you type in the formulas, you get the same answers on your calculators.

This is a step by step (across) way to turn the sine and cosine of 22.5 degrees into fractions with whole number denominators. 

Answers to quiz 9

cosalpha = 1/5_____ sinalpha = 2sqrt(6)/5 _____ tanalpha = 2sqrt(6)

cosbeta = 2/5______ sinbeta = sqrt(21)/5 ______ tanbeta = sqrt(21)/2

cosgamma = 3/5___ singamma = 4/5 _________ tangamma = 4/3

cosdelta = 4/5 _____sindelta = 3/5 ___________ tandelta = 3/4 

cos(alpha + beta) = (2 - 6sqrt(14))/25
sin(alpha + beta) = (sqrt(21) + 4sqrt(6))/25
tan(alpha + beta) = 50(sqrt(21)+sqrt(6))/-500 = -(sqrt(21)+sqrt(6))/10

Corrected from original.  Thanks to Laura Hidrobo.

cos(beta - gamma) = (6 + 4sqrt(21))/25
sin(beta - gamma) = (3sqrt(21) - 8)/25
tan(beta - gamma) = 1 - sqrt(21)/6

Corrected from original.  Thanks to Bi Ting Zhen and Laura Hidrobo.

cos(gamma + delta) = 0
sin(gamma + delta) = 1

tan(gamma + delta) =undefined

The last one is this way because the angles are complementary.

ERROR ON HOMEWORK #12 Corrected in class, but just to make sure.

Cos 72° = (sqrt(5) - 1)/4

Cos 36° = (sqrt(5) + 1)/4

Sorry for the screw-up. See you on Monday.

Half angle formulas

Instead of typing out the half angle formulas, here is some artwork that spells them out nicely.  Tangent is sine divided by cosine, so you would expect the square root sign to be involved, but a little algebraic manipulation makes it go away, as we will see in class Wednesday.

It all depends on whether we multiply the fraction under the square root sign by (1+ cosa)/(1 + cosa) or (1 - cosa)/(1 - cosa).

Here are some practice problems.  Assume the angles are in the first quadrant, which means that half the angle must also be in the first quadrant, so all the major trig functions are positive.

a) cosa = 1/3

b) cosb = 1/4

c) cosc = 1/5

Answers in the comments.

Addition and subtraction of angles and the double angle formula

Addition of angles formulae:               
cos(a + b) = cosacosb   – sinasinb      
sin(a + b) = cosasinb   + sinacosb          

Subtraction of angles formulae:
cos(a – b) = cosacosb   + sinasinb
sin(a – b) = sinacosb  - cosasinb

The double angle formulae:
cos(2a) = cos²a - sin²a
sin(2a) = 2cosasina

For the tangents of the new angles, divide sine by cosine.

Consider two angles a and b, with the following sine and cosine values.

sina = 1/3 and cosa = 2sqrt(2)/3

sinb = 2/3 and cosb = sqrt(5)/3

Find the following values as exact numbers.

sin(a + b)
cos(a + b)
tan(a + b)


sin(a - b)
cos(a - b)
tan(a - b)



sin(2a)
cos(2a)
tan(2a)


Answers in the comments.


 

msinx + ncosx

Adding multiples of sinx and cosx creates a new sine wave that is taller and has a phase shift.  The rules are as follows:

f(x) = msinx + ncosx

Amplitude = sqrt(m² + n²) 

Phase shift: f(x) = 0 when
msinx = -ncosx

With a little algebraic manipulation we get f(x) = 0 when tanx = -n/m. Therefore the phase shift is -arctan(-n/m).  This is where the function will equal 0, but you have to check to see if it is rising at that point or falling.

If it is rising, we can re-write f(x) = sqrt(m² + n²)sin(x - arctan(-n/m)).

If it is falling, f(x) = -sqrt(m² + n²)sin(x - arctan(-n/m)).

I created this picture over at the website WolframAlpha by typing in "plot sinx + cosx and sinx + 2cosx and 2sinx + cosx", without the quotation marks.









This picture is of all the possible choices of +/-sinx +/- cosx.  The inverse tangent of +/- 1 is +/- pi/4, so those are the places where the graphs cross the x-axis closest to the origin.  The new amplitude for all of these is sqrt(1² + 1²) = sqrt(2).  These four examples tell us how the four options work

f(x) = sinx + cosx:  Graph is in blue.  It crosses the axis at x = -pi/4 and goes upwards so it can be re-written as
f(x) = sqrt(2)sin(x + pi/4)

g(x) = -sinx - cosx:  Graph is in green.  It crosses the axis at x =-pi/4 and goes downwards so it can be re-written as
g(x) = -sqrt(2)sin(x + pi/4)


h(x) = sinx - cosx:  Graph is in red.  It crosses the axis at x = pi/4 and goes upwards so it can be re-written as
h(x) = sqrt(2)sin(x - pi/4)

k(x) = -sinx + cosx:  Graph is in yellow.  It crosses the axis at x = pi/4 and goes downwards so it can be re-written as
k(x) = -sqrt(2)sin(x - pi/4)

Rules for the signs in front of the amplitude and the phase shift.

The sign in front of phase shift is positive if m and n agree in sign (either both positive or both negative) and is negative if m and n disagree in sign.

The sign in front of the amplitude agrees with the sign in front of the sine function.  (I incorrectly stated in class it was the sign in front of the cosine function.  Sorry for the error.)

We will have a lab to practice this on Wednesday.
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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, 1½pi, 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.



 





 

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]

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.

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.

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.

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.

"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.

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.

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

Deciding between The Law of Sines and The Law of Cosines.

To completely describe a triangle we should give all three side lengths and all three angles.  If we get the right three pieces of information, we can use either The Law of Sines or The Law of Cosines to find the other three.

Angle-Side-Angle or Side-Angle-Angle (ASA OR SAA):  If we have two angle measurements, the third angle is easily found by subtracting the sum of the two known angles from 180°.  We also have one side length and we know what angle is opposite that length so The Law of Sines is the easiest choice.

Example #1: beta = 30°, gamma = 70°, b = 10"

By subtraction, alpha = (180-30-70)° = 80°, so we can find a and c as follows.

10/sin 30° = a/sin 80°, which becomes sin 80°(10)/sin 30° = 19.69615506...  Let's round this to the nearest thousandth and say a ~= 19.696".


10/sin 30° = c/sin 70°, which becomes sin 70°(10)/sin 30° = 18.79385242...  Let's round this to the nearest thousandth and say c ~= 18.794".

Notice that alpha is the largest angle and a is the longest side.  This is always true that the order of the size of the angles will be the same as the order of the size of the opposite sides.  This helps to check your answers roughly to catch big mistakes like dividing when you should have multiplied or vice versa.
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Example #2: beta = 30°, gamma = 70°, a = 10 m


As before, alpha = 80°, so we can find b and c as follows.


b/sin 30° = 10/sin 80°, which becomes sin 30°(10)/sin 80° =  5.077133059... Let's round this to the nearest thousandth and say b ~= 5.077 m.

10/sin 80° = c/sin 70°, which becomes sin 70°(10)/sin 80° = 9.541888941...  Round this to the nearest thousandth and c ~= 9.542 m.
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Side-Angle-Side (SAS): Find the third side with The Law of Cosines, and then go to the SSS instructions.


Example #3: a = 5', c = 6', beta = 120°

b² =5² + 6² - 2(5)(6)cos120° = 25 + 36 - -30 = 91, so b = sqrt(91) ~= 9.539'
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Side-Side-Side (SSS): rearranging The Law of Cosines, we can get any of the cosines of the angles as combination of a, b and c.

Example #4: a = 5', b = sqrt(91)', c = 6'

We already know that beta is 120° from Example #3, but let's check to see how exact we get it using the Law of Cosines and the inverse cosine function.



cosbeta = (a² + c² - b²)/2ac = (25 + 36 - 91)/2(5)(6) = -30/60 = -1/2

The inverse cosine of -1/2 is exactly 120°.

Now let's find alpha

cosalpha = (b² + c² - a²)/2bc = (91 + 36 - 25)/2(6)sqrt(91) = 102/12sqrt(91) ~= .891042111...

On the calculator, we can [2nd][cos][2nd][(-)] and get 26.9955084.... which is to say alpha ~= 26.996°.

Instead of using The Law of Cosines to find gamma, let's just subtract our two angles from 180° and get gamma = (180-120-26.996)° = 33.004°.

If we use Law of Cosines, the answer is the same to the nearest thousandth of a degree.
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Example #5: a = 5', b = 9.539', c = 6'

We already know that beta is 120° from Example #3, but we are using the approximation of b instead of the exact number.  Let's see if it changes our answer.


cosbeta = (a² + c² - b²)/2ac = (25 + 36 - 9.539²)/2(5)(6) ~= -0.49987535...

The inverse cosine of the answer is 119.9917536...°.  Rounding this to the nearest thousandth of a degree gives us 119.992°, so we are off by a little bit.


Now let's find alpha

cosalpha = (b² + c² - a²)/2bc = (9.539² + 36 - 25)/2(6)(9.539) ~= .891013392...


On the calculator, we can [2nd][cos][2nd][(-)] and get 26.99913319.... which is to say alpha ~= 26.999°.

Instead of using The Law of Cosines to find gamma, subtract the two angles from 180° and get gamma = (180-119.992-26.999)° = 33.009°.

If we use Law of Cosines, the answer is the same to the nearest thousandth of a degree.
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Practice problems. Round inexact answers to a thousandth of a unit.

1) alpha = 75°, beta = 45°, c = 12"

Find a, b and gamma.

2) a = 7 m, b = 8 m, c = 9 m


Find alpha, beta and gamma.


3) alpha = 120°, b = 6', c = 4'

Find a, beta and gamma.

Answers in the comments.

The Law of Cosines

The Law of Sines is a relationship between side lengths of a triangle and the sine of the opposite angle. It is derived from information we get trying to find the height of a triangle relative to one of the sides being called the base.

The Law of Cosines is used to find a third side length when we have two side lengths and the measure of the angle that lies between them, referred to in geometry as SAS or side-angle-side.  At 90°, the cosine is zero and the extra term - 2abcosC becomes 0 and we get the Pythagorean Theorem.

Besides SAS, the Law of Cosines is useful if we know the all three sides (SSS) and want to find the measure of the angles.  By rearranging the variable with algebraic manipulation we get

cosA = (b² + c² - a²)/2bc
cosB = (a² + c² - b²)/2ac
cosC = (a² + b² - c²)/2ab  

The numerator might be 0 or less, but that is not a problem since cosine can take any value from -1 to 1, including 0 when the angle is 90°.  Since no side of a legitimate can be of length 0, we do not have to worry about the denominator being 0, so all the  possible examples we have of triangles will give use three cosine values, and we can use the inverse cosine function to find the three angles.

The Law of Sines and finding the area of a triangle given two side lengths and the angle between them.

 We have several different ways to get the area of a triangle.  The simplest is Area = ½bh,which assumes we know the length of one of the sides and the length of the perpendicular line segment that connects the opposite vertex to the known side length.  If we know all three side lengths, we have Heron's Formula, which we have discussed in previous blog posts and in class.  Let's assume instead we know two side lengths AND the measure of the angle that lies between them.  In the picture above, it is assumed we know the length of side b, and either we know angle A and side c, or we know angle C and side a.  In either case, the height is opposite the known angle in the right triangle formed by the dotted line.  Since sin = opp/hyp we get

sinA = height/c
or
sinC = height/a

This means we have two ways to represent this height,

height = csinA
or
height = asinC

What this gives us is two ways to represent the area that assumes b is the base

Area = ½basinC
or

Area = ½bcsinA

By rearranging the letters we have a similar formula with sin B


Area = ½acsinB

From here, the algebraic manipulation goes as follows.
 

In the pictures I found online the angles are labeled A, B and C.  In class, I usually use alpha, beta and gamma.  The math hasn't changed, just the labels.

An angle in a triangle has to be between 0° and 180°, which means in Quadrant I or Quadrant II, or possibly on the x or y axis.  The value of sine is greater than or equal to zero in that range of angles, so using this method to find the area will always give us a non-negative number.

Examples:  Let's say a = 3 and b = 4.  If we have the measure of angle C, we can use our formula to find the area.  Round the answer to the nearest thousandth.

a) Angle C = 0°
b) Angle C = 10°
c) Angle C = 20°
d) Angle C = 30°
e) Angle C = 40°
f) Angle C = 50°
g) Angle C = 60°
h) Angle C = 70°
i) Angle C = 80°
j) Angle C = 90°

Answers in the comments.

The trig functions all around the unit circle

In this picture, the angle in the first quadrant closes to the x-axis is labeled a°. It is the upper right hand point of the red rectangle.  (90 - a)° is also in the first quadrant and is the upper right hand corner of the blue rectangle. These two angles are complementary.  The other six angles are some "nice" angle away from a° or (90 - a)°, adding either 90°, 180° or 270°.  All the trig values of sine, cosine and tangent for the eight angles labeled here can be derived from the three values sina, cosa and tana.  We already seen the relationship between the trig values of complementary angles.

sin(90 - a)° = cosa
cos(90 - a)° = sina
tan(90 - a)° = 1/tana

When 90° is added, sine and cosine switch values and the cosine of the new angle is negated as follows.

sin(90 + a)° = cosa
cos(90 + a)° = -sina
tan(90 + a)° = -1/tana

90° more than the complementary angle is the same story with those values.


sin(180 - a)° = sina
cos(180 - a)° = -cosa
tan(180 - a)° = -tana

Adding 180° puts us at the antipode, the polar opposite of where we were.  Sine and cosine will just be negated, while tangent will remain exactly the same.


sin(180 + a)° = -sina
cos(180 + a)° = -cosa
tan(180 + a)° = tana

180° beyond the complement is a similar story.


sin(270 - a)° = -cosa
cos(270 - a)° = -sina
tan(270 - a)° = 1/tana

And then we have adding 270°. This is like adding 180° to the values we got when adding 90°.


sin(270 + a)° = -cosa
cos(270 + a)° = sina
tan(270 + a)° = -1/tana

270° more than the complementary angle is the same story with those values.


sin(360 - a)° = -sina
cos(360 - a)° = cosa
tan(360 - a)° = -tana

Example: 60° is one of our "famous" angles.

sin60° = sqrt(3)/2
cos60° = 1/2
tan60° = sqrt(3)

Find the following values using these three pieces of information.

a) sin150°
b) cos240°
c) tan330°
d) tan 120°
e) sin 210°
f) cos 300°

Answers in the comments.

The inverse trigonometric functions.

This week, we have been learning how to find all the trig functions of an angle if you only know sine, cosine or tangent.  Besides that, given any of these values we can find the angle itself by using the inverse trig function buttons.  On the TI-30X II s, the 2nd button is the blue button at the upper left hand corner.  We have already used it for square roots because the square root function is [2nd][x²].  The inverse trig functions of [sin], [cos], and [tan] are [2nd][sin], [2nd][cos] and [2nd][tan], respectively.

Almost none of the "nice" values of sine and cosine, the ones that are rational, give us "nice" angles.  The exception is ½. sin60° = cos30° = ½.

Here are some of the easy fractions for sine and the angles they represent, rounded to as close as the calculator will give them.

If sin alpha = 1/3, alpha  = 19.47122063°
If sin alpha = 2/3, alpha = 41.8103149°
If sin alpha =1/4, alpha = 14.47751219°
If sin alpha =3/4, alpha = 48.59037789°
If sin alpha =1/5, alpha = 11.53695903°
If sin alpha =2/5, alpha = 23.57817848°
If sin alpha =3/5, alpha = 36.86989765°

If sin alpha =4/5, alpha = 53.13010235°

Again, these are not exact.  The actual numbers are irrational.

One useful property of the inverse tangent law is that if we know tan alpha = x, then tan (90° - alpha) = 1/x.  This is to say if we have a line with slope 1/9 and we find the angle measure is beta, then the angle with slope 9 is (90° - beta).  In this particular case the approximate numbers are
tan 6.340191746° = 1/9 and
tan 83.659808254° = 9

Find the following angles for the given values of tangent, rounded to the nearest thousandth of a degree.  Answers in the comments.

a) If tan alpha = 2, alpha  =
b) If tan alpha = 3, alpha  =
c) If tan alpha = 4, alpha  =
d) If tan alpha = 5, alpha  =
e) If tan alpha = 6, alpha  =
f) If tan alpha = 7, alpha  =

Answers to lab 9/14

sin alpha = 1/10

(1/10)² + cos²alpha = 1
1/100 + cos²alpha = 1 
cos²alpha = 1 - 1/100 = (100-1)/100 = 99/100
sqrt(cos²alpha) = sqrt(99/100) = sqrt(99)/10 = 3sqrt(11)/10
cos alpha = 3sqrt(11)/10
tan alpha = sin/cos = 1/3sqrt(11) = sqrt(11)/33

=====


cos beta = 5/7


sin²beta + (5/7)² = 1

sin²beta + 25/49 = 1
sin²beta = 1 - 25/49 = (49-25)/49 = 24/49
sin beta = sqrt(24/49)

sin beta = sqrt(24)/7 = 2sqrt(6)/7
tan beta = sin/cos = 2sqrt(6)/5

=====

tan theta = 3
3²cos theta + cos² theta = 1
9cos² theta + cos² theta = 1
10cos² theta = 1

cos² theta = 1/10
cos theta = sqrt(1/10) = sqrt(10)/10
sin theta = tan theta cos theta = 3sqrt(10)/10


Don't worry if I didn't get around to correcting yours personally.  Everyone who was in class gets credit for the lab.




 
 

Given one trig value for an angle, how to find the other two.

 

We have learned The Trigonometric Identity, 

sin² theta + cos² theta = 1. 

It's a simple application of the Pythagorean Theorem. a² + b² = c² where c = 1. This means if either the sine or cosine of an angle is known, we can find the other value easily enough. Simply square the value you know and subtract that value from 1, then take the square root of the value. This is where our practice with square roots and fractions will come in handy.

Example: sin theta = 1/5

(1/5)² + cos² theta = 1

cos² theta = 1 - 1/25 = 24/25

Taking square roots, we get

cos theta = sqrt(24/25) = sqrt(24)/5 = 2sqrt(6)/5

And, of course, if we know sine and cosine, tangent = sine/cosine.

sin theta = 1/5, cos theta = 2sqrt(6)/5, so tangent = 1/2sqrt(6) or sqrt(6)/12.

sub Example: cos theta = 1/5

It's the exact same work, just finding sine instead of cosine.

sin² theta + (1/5)²   = 1

sin² theta = 1 - 1/25 = 24/25

Taking square roots, we get

sin theta = sqrt(24/25) = sqrt(24)/5 = 2sqrt(6)/5

cos theta = 1/5, sin theta = 2sqrt(6)/5, so tangent = 2sqrt(6).

  

Finding the sine and cosine when we know tangent is different and a little trickier since sin theta/cos theta = tan theta, multiplying by cosine on both sides gives us

sin theta = tan thetacos theta.

We make a substitution into The Trigonometric Identity, do some algebraic manipulation and we can find cosine.  Once we have cosine and tangent, we multiply them together to get sine.

Example: tan theta = 2.

This says that 2²cos² theta + cos² theta = 1

5cos² theta = 1

cos² theta = 1/5

cos theta = sqrt(1/5) = sqrt(5)/5.

multiply cos and tan and we get sin theta = 2sqrt(5)/5.


Here are some practice problems.  Answers are in the comments.


a) sin theta = 1/3

b) cos theta = 1/3

c) tan theta = 1/3

The "famous" trig values between 0 and 90 degrees inclusive

0°
sin 0° = 0
cos 0° = 1
tan 0° = sin0°/cos0° = 0

30°
sin 30° = ½
cos 30° = sqrt(3)/2
tan 30° = sin 30°/cos 30° =sqrt(3)/3

45°
sin 45° =sqrt(2)/2
cos 45° =sqrt(2)/2
tan 45° = sin 45°/cos 45° =1

60°
sin 60° =sqrt(3)/2
cos 60° = ½
tan 60° = sin 60°/cos 60° =sqrt(3)

90°
sin 90° =1
cos 90° =0
tan 90° = sin 90°/cos 90° = UNDEFINED (can't divide by zero)

In the first quadrant, we have these three situations.

As the angle increases, sine increases, from a low of 0 to a high of 1.

As the angle increases, cosine decreases, from a high of 1 to a low of 0.

As the angle increases, tangent increases, from a low of 0 to a high of infinity.

A little preview of coming attractions.  We are not always going to think of sine and cosine as values on the Normal Standard Position Right Triangle, but instead as the x and y values on the unit circle, defined by the formula x² + y² = 1.

When the hypotenuse equals 1.

We know that the trigonometric functions are defined as ratios in a right triangle as follows:

sine (sin) = opposite/hypotenuse

cosine (cos) = adjacent/hypotenuse

tangent (tan) = opposite/adjacent

Consider all the similar right triangles that include the acute angle we call alpha in this picture.  If we think about the triangle where the hypotenuse = 1, our formulas get much simpler.

sin alpha = opposite

cos alpha = adjacent

tan alpha = opposite/adjacent = sin alpha/cos alpha

The tangent of alpha ALWAYS equals sin alpha/cos alpha, but to make sine and cosine equal to the lengths of sides of a triangle, we need the hypotenuse equal to 1.  Here are some other true statements given our picture.

1. sin alpha is the height of this triangle.
2. cos alpha is the base of the triangle, here labeled x.
3. If we have the triangle with the perpendicular lines at the horizontal and vertical and the angle alpha is on the left side, tan alpha is the slope of the hypotenuse.

And here is a statement that is always true for any angle alpha, a statement known as The Trigonometric Identity.

sin² alpha + cos² alpha = 1.

This is a very straightforward application of the Pythagorean Theorem, since all it is says is that

a²/c² + b²/c² = 1.

This is true because when we combine the two fractions, we get

(a² + b²)/c², which is the same as c²/c² = 1.

I say this statement is The Trigonometric Identity.  Other sources will say there are several trigonometric identities.  There are others, but we will see that all of them are just rearrangements of this one.  Make sure you understand this.  Quite often when you do not know the next step in a trig problem, The Trigonometric Identity is the correct path to take.

Practice problems for square roots in the denominator of fractions

Change these fractions with fractions in the denominators and write them in simplest form. Also write them rounded to the nearest thousandth.

a) 1/sqrt(5)

b) 10/sqrt(2)

c) 6/sqrt(12)

d) 5/sqrt(5)

Answers in the comments.

Practice problems for basic trig functions.

In this picture of a right triangle, let's call the hypotenuse c, the short leg opposite angle alpha a, and b will be the long leg opposite beta.  (Another way to say this is a is the height [or altitude, if you prefer], and b is the base.)  Here are some possible values for the lengths.  Given these lengths, find the requested trig functions for the given angles.  Write the values as fractions in lowest terms.

Let a = 8, b = 15 and c = 17.

a) sin alpha = __________

b) tan beta = ___________

c) cos beta = ___________

d) tan alpha = __________

Let a = 7, b = 24 and c = 25.

e) cos alpha = __________

f) tan beta = ___________

g) cos beta = ___________

h) tan alpha = __________

Answers in the comments.

Right triangles and trigonometric functions

Consider the 3-4-5 triangle.  Because 3² + 4² = 5², this is a right triangle. Any triangle similar to this, which means that all the side lengths are increased or decreased by the same scale, is also a right triangle.

This means a 3-4-5 triangle is similar to a 6-8-10 triangle or a 30-40-50 or a .3-.4-.5 triangle. It doesn't matter if we measure these distances in inches or feet or miles or centimeters or meters or kilometers, as long as we are consistent.  In all of these triangles, corresponding angles have NOT changed.  There is the 90° angle and two angles we will call alpha and beta.  We don't currently know the measure of these two angles, but because all three angles have to add up to 180°, alpha + beta = 90°.  Any two angles that add up to 90° are called complementary.


The picture above shows us two ways to look at the right triangle, one from the perspective of the alpha angle and the other from the perspective of beta.  The hypotenuse doesn't change, but the legs are given the labels Opposite and Adjacent.  If we think of the horizontal line segment as the base and the vertical as the height, the horizontal length is Adjacent to alpha and the vertical length is Opposite alpha.

From the point of view of beta, the Opposite and Adjacent are switched, but the Hypotenuse doesn't change.

The three major trigonometric functions, sine, cosine and tangent (sin, cos, tan) can be defined as ratios between the three sides.  Remember that if the angles don't change but we change the lengths, they change by some uniform multiple, so the ratios between the side lengths will remain the same.  Here are the ratios.

Sine = Opposite/Hypotenuse
Cosine = Adjacent/Hypotenuse
Tangent = Opposite/Adjacent

The mnemonic Soh-Cah-Toa is regularly used in classes now to remember the three fractions that define the three functions.  It kind of sounds like something from a Native American language, so the story is that Soh-Cah-Toa was an important chief of some tribe, but that is completely made up.

Another important thing to remember is that if two angles are complementary, that corresponds to the letters co. 

The sine of alpha is the cosine of beta.
The cosine of alpha is the sine of beta.  

 We will later learn about three more trig functions, cotangent, secant and cosecant.  The rule about adding the prefix "co" or getting rid of it will remain the same.


The tangent of alpha is the cotangent of beta.
The cotangent of alpha is the tangent of beta.  

The secant of alpha is the cosecant of beta.
The cosecant of alpha is the secant of beta.  

Heron's formula practice problems.

Here are all the possible triangles with integer side lengths and perimeter = 12, which means s = 12/2 = 6.  Find the areas using Heron's formula, simplifying the square roots.

a) 6, 5, 1

b) 6, 4, 2

c) 6, 3, 3

d) 5, 5, 2

e) 5, 4, 3

f) 4, 4, 4

Answers in the comments.

Two different ways to get the area of a triangle.

If we have three lengths, let's call them u, v and w, we have ways to find out if they could be the three sides of a triangle and if so, we can classify the triangle as isosceles, equilateral or scalene, as well as telling if the triangle is obtuse, right or acute.  With Heron's Formula (sometimes known as Hero's Formula), we can find the area.

Let p be the perimeter, u+v+w.  We will call the semi-perimeter s = ½(u+v+w). The formula for the area is sqrt(s(s-u)(s-v)(s-w)).  The simplest way to do this is to find the semi-perimeter and plug numbers in.

Example #1:  Let the sides be 5, 6 and 7.  The perimeter is 18, so s is 9.  The area is as follows.

sqrt(9(9-5)(9-6)(9-7)) =
sqrt(9 × 4 × 3 × 2) =
6 × sqrt(6) (square units)

Example #2:  Let the sides be 5, 6 and 5.  The perimeter is 16, so s is 8.  The area is as follows.

sqrt(8(8-5)(8-6)(8-5)) =
sqrt(8 × 3 × 2 × 3) =
12 (square units)

This takes more work that A = ½bh, but we aren't always given the height of a triangle.

=======

Here is a completely different way to define a triangle, three points on the xy-plane where one of the points is the origin (0, 0) and the other two are (x1, y1) and (x2, y2).  Now the formula for area is ½|x1y2 - x2y1|.  Let's do some examples of this.

Example #1: (0, 0) (3, -1) and (4, 2)
The area is ½|3×2 - -1×4| = ½|6 - -4| = ½|10| = 5 (square units) 


Example #2: (0, 0) (6, 3) and (4, 1)
The area is ½|6×1 4×3| = ½|6 - 12| = ½|-6| = 3 (square units) 

Practice for slope, distance and area of a triangle defined by three points.

We will deal with the following three points (3, 4), (7, 5) and (-2, 8) as well as the origin (0, 0)

Slope
a) slope from (3, 4) to (7, 5)
b) slope from (3, 4) to (-2, 8)
c) slope from (3, 4) to (0, 0)


d) slope from (7, 5) to (-2, 8)
e) slope from (7, 5) to (0, 0)
f) slope from (-2, 8) to (0, 0)

Distance




g) distance from (3, 4) to (7, 5)

h) distance from (3, 4) to (-2, 8)
i) distance from (3, 4) to (0, 0)


j) distance from (7, 5) to (-2, 8)
k) distance from (7, 5) to (0, 0)
l) distance from (-2, 8) to (0, 0)

Area of triangles
m) area of triangle with points (3, 4), (7, 5) and (0, 0)

n) area of triangle with points (3, 4), (-2, 8) and (0, 0)
o) area of triangle with points (-2, 8), (7, 5) and (0, 0)
p) area of triangle with points (3, 4), (7, 5) and (-2, 8)

 Answers in the comments.
 
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Simplifying square roots and square roots in fractions.

The editing software for the blog does not make it easy to write a square root sign (also known as a radical), so instead I will write sqrt(n) to denote the square root of n.

Square roots of perfect squares

Since 5x5 = 25, which we can also write as 5² = 25, we can simplify sqrt(25) = 5. Any perfect square inside a square root is written more simply as the original number before it was squared then the root was taken, assuming the original number was not negative.

Simplifying the square roots of numbers that are not perfect squares
Consider sqrt(24), which is not a perfect square.  We can find the prime factorization of 24 by a factor tree, a method shown in class, and get that 24 = 2×2×2×3.  If we take sqrt(2×2×2×3), we have a square, 2×2, under the root sign, so the pair may be removed and one is put outside the root sign.  (Here I used the metaphor taught to me so long ago by mean Mrs. Kruger, that the root sign is like a prison, and to escape you need a twin.  One twin escapes and the other is killed, never to be seen again.)  Using the prison break rule.  sqrt(2×2×2×3) = 2×sqrt(2×3) = 2sqrt(6).

Simplifying fractions with square roots
To put a fraction involving a square root in lowest terms, the denominator must be free of radical signs.  For instance, 1/sqrt(n) is not in correct form, so we multiply top and bottom by sqrt(n) and get sqrt(n)/n.

Example: 1/sqrt(8) can be changed to sqrt(8)/8, but 8 = 2×2×2, so sqrt(8) is 2sqrt(2).  Since the 2 in front of the radical is not "in the prison", we can reduce the fraction 2sqrt(2)/8 = sqrt(2)/4.

Practice problems

a) sqrt(50)
b) sqrt(48)
c) sqrt(99)
d) sqrt(98)
e) 1/sqrt(48)
f) sqrt(98)/sqrt(50)

Answers in the comments.

Practice with side lengths and classifications

Here is a list of triples, three numbers in a set.  Determine if the three numbers could be the lengths of triangle sides, and if so, classify the triangle by both classification types,

Classification by biggest angle: Obtuse, Right, Acute
Classification by side length relations: Isosceles, Equilateral, Scalene

a) 6, 3, 3

b) 7, 4, 1

c) 5, 5, 2

d) 5, 4, 3

e) 4, 4, 4

Answers in the comments.

Practice with two angles of a triangle.

In the following problems, two angles of a triangle are given.  Find the missing angle and find the classifications of the triangle based on angle relations (isosceles, equilateral, scalene) and on largest angle (obtuse, right, acute).

1) 70°, 10° ____________ Classifications: ______________ and _____________

2) 70°, 20° ____________ Classifications: ______________ and _____________

3) 70°, 30° ____________ Classifications: ______________ and _____________

4) 70°, 40° ____________ Classifications: ______________ and _____________

5) 70°, 50° ____________ Classifications: ______________ and _____________

Answers in the comments.

Fact about triangles

A triangle is defined by three points, sometimes called vertices (plural of vertex) connected by three line segments, usually called sides.  Two sides meeting at a vertex create an angle, so a triangle also has three interior angles.

The interior angles of a triangle always have a sum of 180°.  This means that if two angles are given, the third can be found using subtraction.

Classification of Triangles.

One classification method for triangles is determined by the largest of the three angles.

If the largest angle is greater than 90°, the triangle is obtuse.


If the largest angle equals 90°, the triangle is right.


If the largest angle is less than 90°, the triangle is acute.

A second method of classification deals with how the angles relate to one another.

If at least two angles have the same measure, the triangle is isosceles.


If all angles have the same measure, the triangle is equilateral. (In this case, all angles are 60°.)

If all angles are different, the triangle is scalene.

An equilateral triangle is a special case of isosceles.

If a triangle is defined by side lengths instead of angles, it is easy to tell which classification is true.

All sides the same length means equilateral.

At least two lengths are the same means isosceles.

All different lengths means scalene.

The Triangle Inequality

If we choose three positive numbers at random, it might be that these cannot be the side lengths of a triangle.  Since these lengths are straight line segments and therefore the shortest distance between two points, the other two lengths have to add up to at least as much as the long length.  (Think of the three points.  If I travel from A to B, it cannot be a short cut if I instead travel from A to C to B.) Likewise, the difference between two distances has to be less than third distance.

If we let the three sides be labeled a, b  and c,  we can state the Triangle Inequality as follows.

a + b > c > |a - b|

If I pick three points at random on a plane, they might all be on the same line.  So if b is directly between a and c, we need to take that in account by changing the greater than signs into greater than or equal signs.
 
a + b >= c >= |a - b|

Some notes about symbols used in trigonometry

Trigonometry is very closely connected to the geometry of triangles, so we will use illustrations like the one above on a regular basis.  Because there are so many different things to consider, we use different alphabets to signify different types of objects.

The standard way to discuss this particular triangle is to call it triangle ABC, where the capital letters in the English alphabet refer to the points.  (A point may also be called a vertex.  The plural is vertices.)  The sides are designated by lowercase English alphabet letters. The side opposite a lettered vertex will share the same letter, only in lowercase.  In this example, side a is opposite point A, side b is opposite point B and side c is opposite point C.

Angles are yet another different object and they are designated with Greek letters.  This blog's editing software is based on HTML, which doesn't have the Greek alphabet symbols readily available, so I will have to type out the words alpha, beta and gamma when referring to angles here.  On any other printed material, the actual Greek letters will be used.  As you can see in this picture, the angle at point A is alpha, the angle at point B is beta and the angle at point C is gamma.

We don't use all the Greek letters angle designations.  One obvious example is pi, which is the Greek letter for p in English and is already being used to mean 3.14159..., the ratio of the circumference of a circle to its diameter.  Besides alpha, beta and gamma, the most common Greek letters used in trigonometry are delta, which looks like a squiggly lowercase d, and  theta,  which looks like the letter o with a horizontal line crossing in the middle.

Syllabus for Math 50 for Fall 2011

Math 50: Trigonometry Fall 2011 – Laney College
Instructor: Matthew Hubbard
MWF: 9:00-9:50 am
Email address: mhubbard@peralta.edu
Recommended Texts: Trigonometry (open source) http://mecmath.net/trig/trigbook.pdf
Barnett et al, "Analytic Trigonometry with Applications" (10th Edition)
Office hours: MWF: 8:30-8:55 am G-210 (Classroom)

Add and drop class dates
Last date to add: Sat., Sept. 3
Last date to drop class without a “W”: Sat., Sept 17
Last date to drop class with a “W”: Wed., Nov. 23

Holiday schedule for MWF schedule
Labor Day: Monday, Sept. 5
Veteran’s Day: Friday, Nov. 11
Day after Thanksgiving Friday, Nov. 25

Test dates:
Midterm 1: Friday, Sept. 23
Midterm 2: Friday, Oct. 28
Comprensive Final: Wednesday, Dec 14 8:00-10:00 am

Homework to be turned in: Assigned last class period of the week, due next class.
Late homework accepted AT THE BEGINNING of the class after it was due

Quizzes: One on the last class of the week in weeks without a midterm or final

Grading system
Homework 20%
Lab 5%
Midterm I 25%*
Midterm II 25%*
Quizzes 25%*
Final 25%

Lowest two of the homework scores will be dropped from the total.
Lowest two of the quiz scores will be dropped from the total.
*Lowest total out of 100 points the quiz total and two midterms will be dropped from the final grade.
Anyone getting a higher grade out of 100 points on the final than the weighted average of all grades combined will get the final percentage instead deciding the final grade.
This option is only available to students who have missed at most three homework assignments.
The extra 5%: Extra credit will be factored into quizzes, midterms and homework at about 5%. This means 100 point tests with have 105 points and homework and quizzes will average 21 points each, though they will count as 20 points.
Class rules:
All cell phones and electronic communication devices off during class.
No hats, hoodies or headphones worn during quizzes and exams.
No calculators that also combine a cell phone or text message machine.

Recommended calculator: TI-30XIIs (any calculator with at least two lines of output will do, the TI-30XIIs is the cheapest that does all the things you need to do in this class. If you need help with any Texas Instruments calculator, I should be able to steer you in the right direction. I haven’t used other brands of calculators as much.)

Open source textbook: The textbook is free and online. It helps a lot if you have a computer and Internet at home so you can have your own copy on your computer. You will need the Adobe Reader software, which is free and can be downloaded from several websites.

Academic honesty: All assignments you turn in, homework, exams and quizzes, must be your own work. Anyone caught cheating on these assignments will be punished, where the punishment can be as severe as failing the class or being put on college wide academic probation.

Student learning outcomes
Math 50 Trigonometry
• Evaluate the 6 trigonometric functions using a calculator, as well as determining exact values for some special angles without a calculator.
• Solve a triangle (right, acute, obtuse), given various angles and sides.
• Convert between decimal degrees, degree-minute-seconds, and radian measure of an angle.
• Demonstrate knowledge of several trigonometric identities and use them to verify other identities.
• Graph trigonometric functions.
• Solve trigonometric equations.