Trig @ MindSay

So I had my first Trigonometry class last night. I find it almost amusing, me saying a few posts back, that I was going to zip right through the course and even double up to take Pre-Calc. Heh. Heh. Very very fucking funny.

 

Anyway, I'll survive, but last night I had to drag out my old college algebra book to refresh how to work with exponential expressions, and I have a feeling that I'll be doing that many more times this term. Still, the feeling that I get when I finally realize that I've solved the problem (or even just figured out how to solve something) doesn't even come close to how I felt learning some new (to me) fact of History. This degree path is seriously going to kick my ass, but I'm going to have fun while I'm spiraling downward, so I guess that's good.

Draw Quadrant I of the XY axes: that is, just draw the top-right portion of how you would normally draw your XY-axes graph. That way you only see the positive x, positive y parts.

Now, on the horizontal axis, going right, draw a line. Where you stop, call this point C. Now draw a vertical line straight up from here, calling this point B where you stop. From the origin (call it point A), draw a ray connecting to point C and going on.

We say the length of AC is x, the length of CB is y, and then the length of AB is r. B has coordinats (x,y).

Now, the side length y is called the side opposite of angle A. This makes sense because y is on the opposite side of A. The side x is called the side adjacent of angle A. This makes sense because the two are right next to each other. Note that for adjacents, you DO NOT LOOK AT THE HYPOTENUSE, r. NEVER. ONLY CONSIDER THE TWO "legs."

Now, from there, we can define our trig functions, which consists of a table you should commit to memory.

sinA = y/r = side opposite / hypotenuse
cosA = x/r = side adjacent / hypotenuse
tanA = y/x = side opposite / side adjacednt
cscA = r/y = hypotenuse / side opposite
secA = r/x = hypotenuse / side adjacent
cotA = x/y = side adjacent / side opposite

Keep that drawing. We'll be using it later on.

Our next table you should commit to memory consists of cofunctions. Cofunctions are complementary, in that the sum of the angles 90^o. And it works like this--

sinA = cos(90^o - A)
cosA = sin(90^o - A)
cscA = sec(90^o - A)
secA = csc(90^o - A)
tanA = cot(90^o - A)
cotA = tan(90^o - A), for all A < 90^o.

How do we gather this? Well, look at our triangle. More specifically, let's look at sinA = y/r.

Now, look at angle B. Look at ITS side adjacent, which is y. We can say cosB = y/r. What else = y/r?

sinA!! Thus, we can say sinA = cosB. Now, we know there is a 90^o angle there called angle C based on our drawing. We know triangles have 180^o. So, we can eliminate that 90^o and get 180^o - 90^o = 90^o.

Thus, A + B = 90^o. Now, what say we subtract A from both sides? A - A + B = 90^o - A, or B = 90^o - A.

Look at that! We just saw how our cofunctions work together! In our table, instead of saying sinA = cosB, we simply replaced B with 90^o - A to get sinA = cos(90^o - A), and we did the SAME THING with every other function there. That's how cofunctions work!

Now, we need to simply analyze two particular types of triangles that come up VERY frequently (so why NOT give them special treatment?). They are dubbed the 30-60 triangle and the 45-45 triangle. We'll start with the 45-45 triangle.

The 45-45 triangle is simply an isosceles triangle with one angle = 90^o and the other two angles = 45^o.

Draw one and just specificy two angles as being 45^o with a 90^o angle there. We know based on how isosceles triangles work that the two sides opposite the two 45^o angles are congruent, or equal in measure, because their angles are congruent as well.

Now, specify any length to the two equal sides.  For simplicity, we'll say that these lengths = 1 each.  Using Pythagorean Theorem, we can, with a = b, say that c² = a² + b² = 1² + 1² = 1 + 1 = 2.  c = Sqrt(2).  We'll denote this "c" as "r", and we'll go ahead and denote the height a (or b) = y and the length b (or a) = x.

Now, taking the triangle to go down, right, up-left (the edges, I mean), this puts the right angle in the lower-left corner.  With x = length = 1 and y = height = 1 and r = hypotenuse = Sqrt(2), we can find the six trigonometric functions for the 45 degree angles.  Take the bottom right 45^o angle as our angle.  We can say that sin(45) = y / r = 1 / Sqrt(2) = Sqrt(2) / 2.  cos(45) = x / r = 1 / Sqrt(2) = Sqrt(2) / 2.  tan(45) = y / x = 1 / 1 = 1.  csc(45) = r / y = Sqrt(2) / 1 = Sqrt(2).  sec(45) = r / x = Sqrt(2) / 1 = Sqrt(2).  cot(45) = x / y = 1 / 1 = 1.

BAM.  We have all 6 functions for ALL angles of 45 degrees.  It does not matter what the lengths of the sides are:  sin(45) is ALWAYS Sqrt(2)/2.  ALWAYS.  Same with the others.

Now, let's look at that 30-60 triangle.  Rather than bother drawing a 30-60 triangle, let's draw an equilateral triangle (all lengths are the same, all angles = 60^o).  We'll let each length = 2.

Now, draw a vertical line to divide this sucker in half, and we'll look at the left half.  As we can easily see, the bottom has been split in half, so no longer is it = 2, but now it = 1.  We'll let this be our length x, and the height = y, and our hypotenuse = r.  Because we split the triangle in half, we also split that top angle in half.  Now, the top angle is 30^o, putting our 60^o down in the lower left corner, and our right angle is in the lower right.  Now, we need to find the length of this height.  By Pythagorean Theorem, we have r² = x² + y².  2² = 1² + y².  4 = 1 + y².  3 = y².  y = Sqrt(3).

Now that we have all three angles and all three sides, we can find the trigonometric functions.  We shall evaluate them at the 60^o angle.

sin(60) = y / r = Sqrt(3) / 2.  cos(60) = x / r = 1 / 2.  tan(60) = y / x = Sqrt(3) / 1 = Sqrt(3).  csc(60) = r / y = 2 / Sqrt(3) = 2Sqrt(3) / 3.  sec(60) = r / x = 2 / 1 = 2.  cot(60) = x / y = 1 / Sqrt(3) = Sqrt(3) / 3.

Now for a little trick.  If you analyze the 30^o angle, you see things are just the opposite.  That is, sin(30) = cos(60) = 1/2.

These are the most fundamental angles, 30, 45, and 60, and they make up something we'll look at later called the Unit Circle.  You better go ahead and start memorizing them because they are angles every aspiring math student should memorize (as you work with them later on in this course and in calculus if you do it, you use these angles so much you really can't help BUT memorize them, even accidentally).

To help, let's make a table!
       sine           cosine       tangent       cotangent   secant   cosecant
30   1/2            Sqrt(3)/2   Sqrt(3)/3       Sqrt(3)      2Sqrt(3)/3       2
45    Sqrt(2)/2   Sqrt(2)/2   1                   1               Sqrt(2)         Sqrt(2)
60   Sqrt(3)/2      1/2          Sqrt(3)          Sqrt(3)/3      2               2Sqrt(3)/3

Now for some problems!

1)  Given a right triangle with x = 20, y = 21, and r = 29, find sinA, cosA, and tanA where angle A is the angle opposite y.

sinA = y / r = 21/29.  cosA = x / r = 20/29.  tanA = y / x = 21/20.

2)  Write cot(73^o), cos(a + 20^o), and tan(25.4^o) in terms of their cofunctions.

Well, we see for the first we have cotA.  cotA = tan(90 - A), so this would be tan(90 - 73) = tan(17^o).
For the second one, it looks ugly with a variable, but don't worry about it.  Just do this normally.  cosA = sin(90 - A) = sin(90 - a + 20) = sin(70^o - a).  The last one is just tanA = cot(90 - A) = cot(90 - 25.4) = cot(64.6^o).

3)  Find one solution for cot(5B + 2^o) = tan(2B + 4^o).

Ew.  Well, just use your cofunction.  We know that cotA = tan(90 - A), so we can rewrite the cotangent part as simply tan(90 - A) = tan(90 - 5B - 2^o) = tan(88^o - 5B).  Putting that = tan(2B + 4^o), the tangents go away, leaving us with 88 - 5B = 2B + 4.  Now it's stupid simple.  84 - 5B = 2B.  84 = 7B.  B = 12.

The first thing we're going to talk about is the reciprocal identity.

Remember how in the last section we said sinθ = y/r, and then we said cscθ = r/y?  Look at that.  y/r, r/y. Why, that's a reciprocal!  THAT'S NO COINCIDENCE!

Let's list 'em!

sin(θ) = 1/csc(θ)
cos(θ) = 1/sec(θ)
tan(θ) = 1/cot(θ)
csc(θ) = 1/sin(θ)
sec(θ) = 1/cos(θ)
cot(θ) = 1/tan(θ)

Notice how we can multiply things to get rid of fractions, too.

1 = sin(θ)csc(θ) = cos(θ)sec(θ) = tan(θ)cot(θ)

In addition to these, all the trig functions follow specific rules when they land in specific quadrants.  Or rather, they have specific signs (+ or -) in specific quadrants.  LISTING TIME!

θ in Quadrant    sinθ   cosθ   tanθ   cotθ   secθ   cscθ
      I                  +        +        +       +        +        +
     II                  +        -         -       -         -         +
     III                 -         -        +       +        -         -
      IV                -         +       -        -         +        -

How do we know that sinθ is ALWAYS negative in QIII?  Well, look at it.  sinθ = y/r, right?  r > 0 for ALL quadrants, so it's ALWAYS positive.  And then y is negative in QIII, so it's a negative divided by a positive, which is a negative.  You can apply the same logic to all the other functions in the other quadrants.

So, if we have cosθ = 2/3, then we know it is either in QI or QIV because the value is positive.  :D

Now, let's talk about the range of these functions.  You can't always just say sinθ will = any number you want.  There are a few rules.  This actually simplifies things, if you think about it.

Look at the triangle we drew last section with x, y, and r segments.  As we rotate the segment r upwards until it is perfectly vertical, we see that the value of y changes, getting larger and larger, but never getting larger than r itself.  It can, however, = r.  So, we say y < r.

If we divide by r on both sides, we see y/r < 1.  If we do the same, only this time we rotate r clockwise so it goes through QIV first, we ultimately get to see that -y < r, which changes to -y/r < 1, which is the same as y/r > -1.

We put the two inequalities together.  -1 < y/r < 1.  What is y/r?  It's SINE!!  So, we can say -1 < sinθ < 1.

...Similarly, we can do this with cosθ, and we have different ideas with secθ and cscθ.  Ultimately, though, these are our ranges:

-1 < sinθ < 1               ,  -1 < cosθ < 1
secθ > 1 or secθ < -1  ,  cscθ > 1 or cscθ < -1
tanθ = anything            ,   cotθ = anything

Now, we only have two more things.  Quotient Identities and Pythagorean Identities.

Quotient Identities.  Let's just randomly decide to throw sinθ / cosθ.  sinθ = y/r, cosθ = x/r.
sinθ/cosθ = (y/r)/(x/r) = (y/r)*(r/x) = y/x = tanθ.  WHOAMG.

We have two of these.

sinθ / cosθ = tanθ, cosθ =/= 0
cosθ / sinθ = cotθ, sinθ =/= 0

Obviously, the denominator cannot be 0 or we get division by 0.  But those are it!  That just leaves...

Pythagorean Identities.  Remeber r² = x² + y².  Let's divide everything by r².

r²/r² = 1 = x²/r² + y²/r² = (x/r)² + (y/r)².  What is x/r?  cosθ!!  What is y/r?  sinθ!!  Put 'em in!!

(cosθ)² + (sinθ)² = 1  -->  cos²θ + sin²θ = 1.

If you do the same process, but instead divide by x² or y², you ultimately get these three Pythagorean Identities (you should go ahead and work out the division by x² and then y² for some practice!):

sin²θ + cos²θ = 1
tan² + 1 = sec²θ
1 + cot²θ = csc²θ

Now for examples!

1)  Find the tangent of cotθ = -3.

Well, tanθ = 1/cotθ, therefore tanθ = 1/-3 = -1/3.

2)  If cscθ = 3, what is sinθ?

Well, sinθ = 1/cscθ, so sinθ = 1/3.

3)  Give the sign of the sine, cosine, and tangent for the angle 406^o.

Well, 406 > 360, so we subtract 360.  406 - 360 = 46^o.  46 lies in QI.  Therefore, all the signs are positive.

4)  Find the value of θ for
tan(3θ - 4^o) = 1/cot(5θ - 8^o).

Well, remember, tanθ = 1/cotθ, which we can just say tanA = 1/cotA.  Since they must be equal, and the trig functions are reciprocals, we ignore EVERYTHING and just look at the stuff on the inside, then set it = each other.  That is,

3θ - 4 = 5θ - 8.  Simplify -->  4 = 2θ.  θ = 2.

5)  Find secθ if tanθ = Sqrt(7)/3, where θ is in QIII.

Okay, we see a secθ and a tanθ in the same thing.  We can't use quotient or reciprocal because the two aren't related at all.  However, they ARE related in the Pythagorean Identity, tan²θ + 1 = sec²θ.

Plug stuff in.  (tanθ)² + 1 = (secθ)².  Secθ = Sqrt(7)/3, (secθ)² = (Sqrt(7)/3)² = 7/9.

So, we have (tanθ)² + 1 = 7/9.  Subtract 1, or 9/9, from each side.  (tanθ)² = (7/9) - (9/9) = -2/9.

So, (tanθ)² = -2/9.  We can't take the Sqrt of this, so let's just use the positive value for now (it's okay, as we'll learn later.  Ultimately, if you get a negative and need to take a square root, just turn it positive).  Sqrt(2/9) = Sqrt(2)/3.

tanθ = Sqrt(2)/3.  Now, we simply use a calculator to find the angle of it by hitting 2nd TAN (on a TI calculator) to get tan^-1(tanθ) = tan¹(Sqrt(2)/3).  tan¹(tan) cancels out, so we have θ = tan^-1(Sqrt(2)/3) = 25.23940182.

However, we were told to find it in QIII.  To do this, we simply add 180 degrees to this.  25.239 + 180 = 205.239^o.


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What if we were asked to find QI?  Well, the 25 < 90, so it's already in QI.  What if QII?  We just do 180 - number.  What if QIV?  We do 0 - number.  Keep that in mind for other problems.


Why do we care?  ALL those rules help simplify crap super quickly when we need to, and they help us reach answers we otherwise couldn't reach because, really, we just wouldn't know.  However, the most important things are the Pythagorean Identities.  MEMORIZE THEM.  They come up all the time in later math classes in helping to reduce a LOT of crap (especially when you see a sin²θ + cos²θ under a square root sign, and if you know that that = 1, then you just take the square root of 1, which is 1.  VERY handy).

Now we should actually introduce the functions we'll be using from here on out.  They are, of course, trig functions.  They consist of the sine, cosine, trangent, secant, cosecant, and cotangent.

If we draw a point on our origin (0,0), then move to the right some x distance and move up some y distance, then draw a line segment r from the y's endpoint down to the origin, we have a triangle.  We can measure the length of r using the distance formula, r = Sqrt((x - x₀)² + (y - y₀)²), where x and y are arbitrary distances, x₀ and y₀ are the original points of (x,y), which in our case is the origin, and r is the length of the hypotenuse.

Because x₀ = y₀ = 0, we can simplify the above to Sqrt[(x-0)² + (y-0)²] = Sqrt(x² + y²).

Now, so we have the values of x, y, and r.  Now we can define the 6 trig functions.  First of all, looking at this triangle, the little angle between the x segment (the bottom, horizontal segment) and the diagonal slant (r) makes an angle we will just call θ.

Now, let's define.

Sine:  sin(θ) = y/r, r =/= 0.     =/=  means "is not equal to."
Cosine:  cos(θ) = x/r, r =/= 0.
Tangent:  tan(θ) = y/x, x =/= 0.
Secant:  sec(θ) = r/x, x =/= 0.
Cosecant:  csc(θ) = r/y, y =/= 0.
Cotangent:  cot(θ) = x/y, y =/= 0.

Obviously, we can't have 0 in the denominator or else that's division by 0, which is bad.

Also, note something about the tangent, tan(θ) = y/x.  Haven't you seen y/x somewhere before?  Aha!  Slope!  IN GENERAL, tan(θ) = m, so the tangent is the same as the slope.  Not ALWAYS, but USUALLY.

Now for a simple table before we do some examples.  Remember quadrantal angles?  Those things on the XY axes themselves?  0^o, 90^o, 180^o, 270^o, 360^o?

It's good to just memorize this table (you ultimately, as you work in Trig, just come to naturally memorize them because you will use them a lot.  As well as something called the Unit Circle, which we will tackle later).

Anyway, here's our table:

θ       sinθ    cosθ    tanθ           cotθ       secθ        cscθ
0        0        1         0          Undefined     1        Undefined
90      1        0      Undefined      0        Undefined    1
180    0        -1         0         Undefined    -1       Undefined
270   -1        0      Undefined     0        Undefined    -1
360    0         1         0         Undefined     1         Undefined

Also, a huge note:  UNTIL WE HIT CHAPTER 3, KEEP YOUR CALCULATOR IN DEGREE MODE.  For TI calculators, just hit the Mode button, then scroll until you see the row that says Radians or Degrees, and highlight Degrees and hit Enter, then 2nd Mode to Quit.

Now for some examples!!

1)  Find the 6 trigonometric values for the point (x,y) = (0,2).  Rationalize the denominator if applicable.

Okay, we have x = 0, y = 2.  Now we need r.  Remember the first thing?  r = Sqrt(x² + y²) = Sqrt(0² + 2²) = Sqrt(0 + 4) = Sqrt(4) = 2.  You may think you need +2, but you are wrong.  In this case, our "r" is a "distance," which is ALWAYS measured with positive values.  Later, we will see negative r's, but so long as you are taking square roots of things, you stick to positive values.

So, x = 0, y = 2, r = 2.  Now we just do the trig functions:

sin(θ) = y/r = 2/2 = 1.
cos(θ) = x/r = 0/2 = 0.
tan(θ) = y/x = 2/0 = undefined.
csc(θ) = r/y = 2/2 = 1.
sec(θ) = r/x = 2/0 = undefined.
cot(θ) = x/y = 0/2 = 0.

Bam, done.

2)  Given that the quadrant your point (x,y) is in quadrant IV, decide if x/y is positive or negative.

Well, quadrant IV is the bottom-right.  This means x = positive (right), y = negative (bottom).  So, we have +/-, which is negative.

3)  Use the table of quadrantal angles to evaluate the expression:  tan360^o + 4sin180^o + 5cos²180^o.

According to our table, tan360^o = 0, sin180^o = 0, and cos180^o = -1.  For sin180, we multiply this by 4 because we have 4sin180, so 4*0 = 0.  For the cos², that is the same thing as writing cos(θ)cos(θ).  So, we know cos180 = -1, and we just multiply that by cos180, which is also -1.  -1 * -1 = +1.  Then we multiply that by 5.  5 * 1 = 5.

Now put them all together.  0 + 0 + 5 = 5.

4)  If n is an integer (a non-decimal, non-fraction number including negatives, 0, and positives), n*180^o represents an integer multiple of 180 degrees, and (2n + 1)*90 degrees represents and ODD integer expression.  Decide if the expression is = 0, -1, 1, or undefined;  cos[(2n + 1)*90].  (note:  Mindsay has failed and refuses to let me make the degree symbol anymore for this entry.  As a result, assume the 90 is 90 degrees, while the rest are just 2n + 1).

Okay, we have cos[(2n + 1) * 90 degrees].  Look at our table for what cos(90) is.  It is ALWAYS 0.  Now, no matter WHAT integer we put in for n, we will ALWAYS get an odd number out.  2*any integer is always an even number, right?  Add 1 to it, and you ALWAYS get an odd.  So, we ALWAYS get an odd * 90 degrees.

What is an odd*90 degrees.  Let's try some odd numbers, like 1.  1*90 = 90.  3.  3*90 = 270.  5*90 = 450, and so on.  Note that 450 is > 360, so we should do 450 - 360 = 90 to get an idea of where this value is.

If we do it infinitely, we constantly alternate between 90 and 270 degrees.  The cosine at EITHER of these angles is ALWAYS 0.  Thus, our answer is 0.

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Why do we care?  Dude, I'm not really even going to try.  You need to know the 6 trig functions because, hello, this is TRIGONOMETRY.  We will be using them EXTENSIVELY, and about NONSTOP pretty soon.  Best to learn to love them.

We gotta define some words here.  Don't worry.  They're simple.

Draw two points.  Label one A and the other B.  Draw a line through them, and draw arrow heads at the end of each line.  This is called line AB.  The portion consisting ONLY of A and B and everything in between is called a line segment, in this case, segment AB.  Take only point A, draw a line STARTING from A and hits B then goes THROUGH B, with an arrow head on the end.  This is called ray AB.

A line has infinite length and NO endpoints.  A line segment has finite length, the distance from A to B and has TWO endpoints.  A ray has ONE endpoint and infinite length.

An angle is formed by rotating a ray about its one endpoint, in our case, A.  You can also think of it as two distinct rays that go off in different direction, and the angle is the measurement BETWEEN those two rays.  If you rotate the ray counterclockwise, the angle measurement is positive.  If you rotate it clockwise, its measurement is negative.  If you rotated a ray 30 degrees counterclockwise, you have an angle of 30 degrees.  If you rotated it 30 degrees clockwise, you have an angle of -30 degrees.  Got that?  We use denote these as n^o for degrees, where n is any number.  18, 90, whatever.  The important thing is that by tacking the little o thing, we are being told it is a degree.

Now, the ray's initial position, almost always horizontally and pointing to the right because that's the convenience most mathematicians and textbooks use, is called the initial side.  When you rotate it, the new position is called the terminal side.  The endpoint of the ray is called a vertex.

A measurement of an angle, for right now, is called a degree (later, we will discover radians).  You probably know this already, though.  If the angle < 90 degrees, it is called an acute angle.  If = 90 degrees, it is called a right angle.  If > 90 degrees, it is called an obtuse angle.  If exactly 180 degrees, it is called a straight angle.

If you sum two angles and you get exactly 90 degrees, the two angles are complementary angles.  Say, 45 degrees + 45 degrees.  Or 30 + 60.  Or 80 + 10.  Whatever.  If the two angles sum to be 180 degrees, they are supplementary angles.  179 + 1.  Two right angles (90 + 90).  Whatever.

Now, a degree isn't the only way we measure angles.  Sometimes we need decimals in our angles.  What if you have 20.787878 degrees?  Well, can you actually visualize that?  Probably not.  Actually, you probably won't be able to visualize these next two terms, but you need them for calculation and conversion purposes.

The first "decimal measure" is called the minute.  We denote this as n', where n is just some number. ANY number.  The important thing is that we use ONE apostrophe to tell us it is a minute.  Do NOT confuse this with the TIME "minute," which is 60 seconds.  The GEOMETRIC minute is simply 1/60 of a degree.  If you have 2 degrees, you have 2 / (1/60) = 120 minutes.  :D  It takes 60 minutes to make 1 degree, so, to convert, you simply do

x degrees   *   60 minutes
      1                1 degree

The word "degrees" cancels out (we are actually cancelling units), and the only word that remains is "minutes."  Likewise, to convert from minutes to degrees--

x minutes  *  1 degree
     1             60 minutes

The next one is called a second.  DO NOT CONFUSE WITH THE TIME SECOND.  A second is 1/60 of a minute.  We denote this by n'', where n is some number, ANY number.  We use TWO apostrophes here.  Our conversions are--

x degrees   *   60 minutes   *   60 seconds
    1                   1 degree           1 minute

x seconds  *   1 minute     *   1 degree
    1                60 seconds    60 seconds

If you only need to go from seconds to minutes, just drop the 1 degree / 60 seconds on the end of that second one there.

Remember when I said that most mathematicians and textbooks have the rays pointing to the right?  There's a term for this.

If, on a graph, the vertex is at the origin (0,0) and the ray points perfectly straight to the right along the x-axis, we call this the standard position.

We then rotate this angle throughout the four quadrants of the graph.  Quadrant I is the top-right part of the graph.  Quadrant II is the top-left.  Quadrant III is the bottom-left.  Quadrant IV is the bottom-right.  KNOW THIS.  See your actual axes?  The XY axes?  Those "lines" actually have specific degree measurements.  The positive-y (pointing up) is 90 degrees.  Negative-x (pointing left) is 180 degrees.  Negative-y (pointing down) is 270 degrees.  And positive-x (pointing right) is 0 or 360 degrees.  Each of these four (actually five if you count both 0 AND 360) "angles" is called quadrantal angles.  This is because they DIVIDE the four quadrants.

If you make a COMPLETE rotation, that is, go a full 360 degrees so you're back in the standard position, you are a coterminal angle.  ALL angles > 360 have at least ONE coterminal angle.  To find just where the ray points if it's > 360, simply subtract 360 from the angle.  For example...

Suppose we have an angle of 450 degrees.  What is this?  Well, let's subtract 360 from 450.  450 - 360 = 90.  Bam.  Our ray now falls on the 90 degree angle!!  See how easy that was?

Let's do some examples!!

1)  62^o 18' + 21^o 41'.

All we're doing is adding things SEPARATELY.  Remember, a minute is NOT a degree.  It's a PART of a degree, so you can't just carry numbers over together.  Let's add the minutes first.  41 + 18 = 59.  That's not 60, so we're okay.  Now let's add the degrees.  21 + 62 = 83.  So, our final answer is 83^o 59'.

2)  180^o - 124^o 51'.

We don't have any minutes with the 180.  However, we remember that one degree is 60 minutes.  So, why don't we take off one degree so we can have 60 minutes?  180 degrees - 1 degree = 179 degrees.  Now we have these 60 minutes in there, too.  Put them together!  179^o 60'.  Remember, if you don't have any units you need, just take one away from the units you HAVE to get 60 of the units you need (assuming what you need is smaller than what you have.  Minutes are smaller than degrees).

Now instead we have 179^o 60' - 124^o 51'.  Now we can subtract!  Subtract separately, now.  179 - 124 = 55.  60 - 51 = 9.  Our final answer is 55^o 9'.

3)  Find the angle of smallest measure coterminal with the angle -40^o.

What does this mean?  We know the coterminal angles are 0, 90, 180, 270, and 360.  -40 is in Quadrant IV, so we know it has to be either 0 or -90 (-90 is the same as 270.  We measured counterclockwise for being positive, so we can measure clockwise for negative).  Well, -40 is closer to 0 than it is to -90, isn't it?  So, our answer is 0^o.

4)  Convert 31.4296^o to degrees, minutes, and seconds.

Well, we know we can just take the 31^o by itself.  Now we're left with .4296^o we need to convert.  Why?  Because minutes and seconds are FRACTIONS of a degree, that is, they are < 1 degree.  This decimal < 1, right?  So, yeah, we only worry about it for degrees.

If we remember, to get from degrees to minutes, we multiply by 60 (scroll up).  So.. 0.4296 * 60 = 25.776.  Now we can hack off the 25 because it's some minutes, leaving the decimal to be converted into seconds (because a second is a FRACTION of a minute, just like what we did a second ago).  Now, to convert from minutes to seconds, we multiply by 60 again.  0.776 * 60 = 46.56.

Let's put everything together.  31^o 25' 46.56".

5)  A pulley rotates through 75^o in 1 minute.  How many rotations does it make in 1 hour?

A word problem!  We have 75^o per 1 minute, and we have 1 hour.  Let's do some CONVERSIONS!

75 degrees     *    60 minutes    *   1 revolution   =  12.5 revolutions/hour.
    1 minute                1 hour           360 degrees

Whoah, how did I do all that?  You start with what you're given.  We were told we have 75 degrees in one minute.  That also means per minute.  So, we have our first fraction.  We know there are 60 minutes in 1 hour, so we set up the next fraction as a way to "cancel out" the word "minutes."  It WAS in the denominator, so we put it in the numerator because anything / itself = 1.  We did the same thing with the degrees in the NEXT one (though, you could have done this one first).  360 degrees are in 1 revolution.  And then it's just multiplying.

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Why do we need to know about angles?  Well, geez, why not?  If you don't know anything about angles, you can't do the real trig stuff later LOLOLOL.  Also, the key point here was how angles work.  They are measured ROTATIONS about the vertex.  CD players rely on this.  CD players rotate CDs at a very specific speed.  You need to know this speed in order to determine how much information CDs can store and at what layering so the CD player's speed can read it correctly and not skip or scratch the disc.