Question

Trigonometry:
In the circle, each of the line segments listed below corresponds to one of the six trigonometric functions of angle, theta. Determine which one matches each line segment: PN, PM, AR, BQ, OR, and oQ and show how you came to those answers. Do not assume numbers or angles not given. Do not put specific numbers on angles or points unless given.

Answer 1

Please! Help! photo attach

Answer 2

PM is sine and PN is cosine. This comes from the fact that sin is related to the leg of the interior triangle opposite the angle theta and cosine is related to the adjacent leg of the triangle.
AR is the tangent and BQ is the cotanget (notice that both of these lines are tangent to the circle). The tangent is always the line parallel to the leg representing sin and intersecting the extended hypotenuse of the inner triangle, and the cotangent is similarly parallel to the leg representing the cosine and intersecting the hypotenuse.
The length of the extended hypotenuse where it intersects the tangent line is the secant and the length where it intersects the cotangent line is the cosecant, so 0R is the secant line and 0Q the cosecant.

Answer 3

Just remember that the 3 co- functions are related and the 3 non-co functions are related. The cosine is adjacent to theta, the cotangent is parallel to this starting at the y axis and going until it intersects the extended hypotenuse, and the cosecant is the length of the hypotenuse where this intersection occurs, while the 3 non-co functions share the same relationship but sine is the opposite side of the triangle from theta.

Answer 4

Nice very well done! i can't believe i spend hours trying to figure this one out

Answer 5

do you mind taking a look at another problem online?

Answer 6

Not at all

Answer 7

http://www.oglethorpe.edu/faculty/~j_nardo/Academic%20Talks/GTMC%202000/GA%20Math%20Conference%202000.htm

Answer 8

part 5 or part V is the one i need help understanding. can you explain that to me?

Answer 9

Sure, I can give it a shot.

Answer 10

Thanks for helping me by the way

Answer 11

We're using trigonometry here to perform a simple population model. We see that in December our population hits a minimum of ~5000 and in June it hits a maximum of ~15000. Since sine functions model things rising and falling periodically, we want to use a sinusoidal function here to model these changes.

Such a function will be of the form p = a + b*sin(mx + c)
where p is the population at month x, a is the average value of the population, b is the variation in population, m is the coefficient of period, and c is the shift in period.

Sine has a maximum value of 1 and a minimum value of -1, so at its maximum and minimum points it will vary an equal amount from your starting value. Thus, we want to know what the average value of the population is. We find this by just taking the average of the minimum and maximum populations:
(5000 + 15000)/2 = 10000
So then our starting value, the value the function should take on when the sine part of the equation = 0, should be 10000.
Thus, a =1000

Next we need to know what the variance is. Since sine has a maximum value of 1 and a minimum of -1, we need to multiply the term times a coefficient to get it to vary as much as we want it to. Since the minimum and maximum values of our population are 5000 away from the center, we want this coefficient to be 5000. This way, when sine takes on its maximum value of 1, the value of the entire term will be 5000 and similarly at the minimum value for sine the term will be -5000.
Thus, b = 5000

Now, sine has a maximum value of 1 occurring at pi/2 and every 2pi thereafter, so pi/2, 5pi/2, 9pi/2, etc. Sine has a minimum value of -1 occurring at 3pi/2 and every 2pi thereafter, so 3pi/2, 7pi/2, 11pi/2, etc. We're using time in months here, with December = 0 as our base point, so the month will be June at x = 6, 18, 30, etc and December at x = 0, 12, 24, 36, etc. Thus, we want our minimum population to occur at x = 0 + 12k (for all integers k) and our maximum to occur at x = 6 + 12k.

Since sine goes through 1 full cycle of 0 > max > 0 > min > 0 every 2pi, this means that we want our input at x = k + 12 to be 2pi greater than our input at x = k. Thus, we want out coefficient b for x to be such that b*x + 2pi = b*(x+12). Solving this equation for b gives us:
b*x + 2pi = b*(x+12)
bx + 2pi = bx + 12b
2pi = 12b
b = pi/6
So we know the coefficient for x inside the sine is going to be pi/6.
Thus, m = pi/6.

Now, as previously mentioned we want the the minimum of the function to occur at x = 0 + 12k and the maximum to occur at 6 + 12k. This is a problem because ordinarily sine takes on its average value of 0 at x = 0. Thus, we need to add a shift term so that when x = 0, the term inside the sine is such that sin will take on its minimum value. Since sine takes on its minimum value at x = 3pi/2, we will use this as our shift. (note that the notes you linked to use -pi/2 - since sine takes on its minimum value at 3pi/2 + 2*pi*k, the value of sine at -pi/2 is the same as the value at 3pi/2)
Thus, c = 3pi/2

Then your final equation will be
p = 10000 + 5000*sin(pi/6*x + 3pi/2)
or, as used in the notes you linked,
p = 10000 + 5000*sin(pi/6*x - pi/2)

Answer 12

No problem, happy to help. I hope that tells you everything you're looking for, but i know it's a lot of material, so if you have any questions post and I'll try to answer them.

Answer 13

By the way, that should be a = 10000

Answer 14

i am picturing the graph would look like this. on the notes theres no number or points?
does this look right to you? where would you put the other tick marks or point?

Answer 15

Obviously the graph of a sine function is a bit less pointy than that, but yeah, that looks about right. Bear in mind though, that when x = 0, the function should be at a minimum (and so the value should be 5000). Then the function should intersect the line y = 10000 at x = 3, 9, 15, 21, etc, and should have minimums at x = 0, 12, 24, 36, ... and maximums at x = 6, 18, 30, 42, ...

Answer 16

thanks im just not really good at drawing. =) should i use number like 1, 2, 3, 4, 5, 6... for my tick marks on the graph? or use those pi/2

Answer 17

Well, x is going to take on values of 1, 2, 3, 4,... The value inside the sine function will take on values of multiples and fractions of pi, but since the x axis is a measure of the value of x, not the value of the input into sine, you should use 1, 2, 3, 4,...

Answer 18

so how would i calculate, how many rabbits population are there in October?

Answer 19

do i just plug in 10 for x?

Answer 20

Exactly!

Answer 21

thanks i'll figure out the graph. =)

Answer 22

1 more question. =) on the note link the found the phase shift by..

"Lastly, the graph must be shifted to the right in order to match the data. Our sine function has a maximum at x = 3 (March); the maximum temperature in Savannah occurs at x = 6 (June). We must shift to the right by 3 units."

why is the sine function has a maximum at x = 3 (march) and maximum temp at x=6
is this a typo?

Answer 23

x = 6 (june)

Answer 24

No, it's just kind of weirdly phrased. What it's saying is that if they just use the period coefficient they found with no shift, the sine function's maximum occurs at x = 3 (in much the same way that in the population example, without adding -pi/2 to the interior of the sine, our maximum would occur at 3 even though the actual maximum from the data is at 6).
Since the actual maximum is at x = 6, there needs to be a shifting factor added into the equation to accommodate this.

Answer 25

thanks a lot for your help. i really appreciated.

Answer 26

Any time!

Answer 27

hello blacksteel! i appreciate your help the other day. i just wanna ask you "are you 100% sure that your answer to the first original question is correct? i have to turn it in and i need the grade. thanks again!

Answer 28

Hey! nevermind! its right!