anonymous
  • anonymous
For an angle Θ with the point (−20, −21) on its terminating side, what is the value of cosine?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@kelliegirl33 can you help me
myininaya
  • myininaya
If it helps you graph the point. Can you do that for me first?
anonymous
  • anonymous
@myininaya just plot -20,-21 ?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

myininaya
  • myininaya
Yep plot the point as a first step.
myininaya
  • myininaya
You can use the draw tool.
anonymous
  • anonymous
ok
anonymous
  • anonymous
|dw:1403474666905:dw|
anonymous
  • anonymous
@myininaya
myininaya
  • myininaya
|dw:1403474755600:dw| Now I will simply make a right triangle using the center of the graph for my central angle.
myininaya
  • myininaya
what is the ratio definition of cosine of an angle?
anonymous
  • anonymous
legnth of adjacent / legnth of hyp
myininaya
  • myininaya
right you already have the adjacent side given in the picture the only thing you need now is the hyp of that triangle, right?
anonymous
  • anonymous
yes
myininaya
  • myininaya
Find the hyp of that triangle using Pythagorean Theorem.
anonymous
  • anonymous
the adjacent side is -20
myininaya
  • myininaya
You are correct to say that.
myininaya
  • myininaya
Can you find the length of the hyp?
anonymous
  • anonymous
yes hold on wouldnt it be -20^2 + 21^2 = c^2
myininaya
  • myininaya
20^2+21^2=c^2 and then take the square root of both sides to give you c.
anonymous
  • anonymous
29
myininaya
  • myininaya
so that is the length of the hyp so what is cos(theta)=?
anonymous
  • anonymous
29 ?
myininaya
  • myininaya
you said earlier cos(theta) is adj/hyp you never said cos(theta) is hyp
anonymous
  • anonymous
20/29
myininaya
  • myininaya
don't forget the negative sign
anonymous
  • anonymous
thank you so much i have a few more cn yu help me
myininaya
  • myininaya
if the angle was on the side of the graph like on the right hand side of the y-axis cos would have been indeed a positive number but since the angle was to the left of the y-axis cos will be negative
myininaya
  • myininaya
|dw:1403475467323:dw|
anonymous
  • anonymous
How does changing the function from f(x) = 3 sin 2x to g(x) = 3 sin 2x + 5 affect the range of the function?
myininaya
  • myininaya
we could even talk about sine sin is positive when above the x-axis and sine is negative when below the x-axis |dw:1403475522177:dw|
myininaya
  • myininaya
Well you are talking about shifting the graph up when you have that +5
anonymous
  • anonymous
so the function shifts up 5 units so the range changes from -1 to 1
anonymous
  • anonymous
in f(x) to 4 to 6 in g(x)
myininaya
  • myininaya
For example, pretend we have y=sin(x) the range of sin(x) is from -1 to 1 (inclusive). but when we look at y=sin(x)+1 that means you are moving every point of y=sin(x) up one unit so the range of sin(x)+1 is from -1+1..1+1( inclusive) doing the addition gives 0..2 (inclusive)
anonymous
  • anonymous
am i right
myininaya
  • myininaya
well first what is the range of y=3sin(x)?
anonymous
  • anonymous
f(x) = 3 sin 2x to g(x) = 3 sin 2x + 5 affect the range of the function
myininaya
  • myininaya
yes the range of the graph of f would be affected if you need f+5
myininaya
  • myininaya
did f+5 (not need)
myininaya
  • myininaya
but the range of y=3sin(x) is not -1 to 1.
anonymous
  • anonymous
Which of the following functions has the greatest y-intercept?|dw:1403475803380:dw|
anonymous
  • anonymous
its 2 to 8
myininaya
  • myininaya
Let y=sin(x). Let f(x)=3sin(x). Therefore you could say f(x)=3y where y=sin(x). If you have 3 times every y per x wouldn't the range be stretched more?
anonymous
  • anonymous
yes
myininaya
  • myininaya
like the range for y=sin(x) is -1 to 1 (when we talk about range we are talking about the y values) the lowest y value for y=3sin(x) would be what then?
anonymous
  • anonymous
so f(x) has greater y-intercerpt
myininaya
  • myininaya
the lowest y value for y=sin(x) is -1 so the lowest y value for f(x)=3sin(x) is ?
anonymous
  • anonymous
-3
myininaya
  • myininaya
great so it shouldn't be hard for you to see the highest y-value for f(x)=3sin(x) would be 3(1)=3 so the range for y=3sin(x) is -3 to 3 (inclusive)
myininaya
  • myininaya
now what is the range for y=3sin(x)+5
anonymous
  • anonymous
2 to 8
myininaya
  • myininaya
omg you are great so now you can answer that one question about how the range is affected and actually state the ranges of both functions
anonymous
  • anonymous
now for Which of the following functions has the greatest y-intercept? wouldnt it be f(x) since the y is 9
myininaya
  • myininaya
Also when answering the question it might be better to choose to answer in interval notation or inequality notation. 2 to 8 isn't really something people say to state an interval (not that I have ever seen anyways)
anonymous
  • anonymous
Okay , so it would be stated has 2 to 8 in g(x)
myininaya
  • myininaya
well like i said you might want to choose to answer in interval notation (or inequality) y is an element of the interval [2,8] <--interval (i put brackets to mean to included endpoints inequality notation: \[2 \le y \le 8\] and to your next question.... |dw:1403476409004:dw| do you what the y-intercept is for a curve? It is the part of the curve that goes through the y-axis. When I say part, I'm talking about the specific point in which the curve does that. What do you know about all the points on the y-axis? What do they all have in common?
myininaya
  • myininaya
I will give you a hint. All the x values of any point on the y-axis are _____?___
anonymous
  • anonymous
Im not sure
anonymous
  • anonymous
4-9 = -5 which is the same for g(x)
anonymous
  • anonymous
the same y-intercept
myininaya
  • myininaya
Ok since you can't answer that, I will ask you a series of questions now? Where does (0,10) lie? Where does (0,5) lie? Where does (0,2) lie? Where does (0,0) lie? Where does (0,-1) lie? Where does (0,-6) lie? where does (0,-1000) lie?
anonymous
  • anonymous
on the x axis
myininaya
  • myininaya
That is incorrect. They all lie on the y-axis. The answer to my question: All the x-values of any point on the y-axis is ____?__ All the x-values of any point on the y-axis is zero. So to find the y-intercept for a curve you just need to find y when x is zero. You just need to replace x with zero and solve for y.
anonymous
  • anonymous
so 0-9 ? 0-4 ?
myininaya
  • myininaya
For the one function, it should be clear (0,0) is the y-intercept (that is the origin of the graph after all; is is all an x-intercept) The other function you did right even though it doesn't seem you know why you were doing it since you were not able to answer the question.
anonymous
  • anonymous
Yeah im confused now
myininaya
  • myininaya
|dw:1403477105487:dw| This is the information given. You are asking to find the y-intercept for both graphs right? Can you tell me which point from f have the x is 0?
anonymous
  • anonymous
functions has the greatest y-intercept
anonymous
  • anonymous
1,1 ?
myininaya
  • myininaya
I'm asking you to tell which of the points from f have the x value is 0?
myininaya
  • myininaya
|dw:1403477214481:dw|
anonymous
  • anonymous
its 0,0 ?
myininaya
  • myininaya
|dw:1403477221682:dw| Do you see any of these x's being 0?
anonymous
  • anonymous
yes
myininaya
  • myininaya
yes the point (0,0) is a y-intercept because the x is 0
myininaya
  • myininaya
now we need to look at g(x)=5cos(3x)-5
myininaya
  • myininaya
what point on g gives us the x is 0?
anonymous
  • anonymous
-5
myininaya
  • myininaya
how did you get that?
myininaya
  • myininaya
Did you replace the x with 0 to see what y is? Because y is not -5 when x is 0.
myininaya
  • myininaya
I only care about when x is 0 since you are asking about the y-intercept.
myininaya
  • myininaya
Replace the x with 0 like so y-intercept of g is g(0)=5cos(3*0)-5
myininaya
  • myininaya
evaluate 5cos(3*0)-5
anonymous
  • anonymous
5cos(0)-5 5cos
myininaya
  • myininaya
you are right to say 5cos(3*0)-5 can be wrriten as 5cos(0)-5 but where do we go from there?
anonymous
  • anonymous
1.41
myininaya
  • myininaya
:( 5cos(0)-5 what is cos(0)?
anonymous
  • anonymous
1
myininaya
  • myininaya
5(1)-5 can you simplify from here now?
anonymous
  • anonymous
0
anonymous
  • anonymous
same y
myininaya
  • myininaya
right so the y-intercept for g is (0,0) as well they have the same y-intercept
anonymous
  • anonymous
1f sin Θ = 15 over 17, use the Pythagorean Identity to find cos Θ.
myininaya
  • myininaya
draw a right triangle using this information sin(theta)=15/17
anonymous
  • anonymous
\[\sin^2 \theta + \cos ^2\theta = 1\] will be what we use ?
myininaya
  • myininaya
you can use that
anonymous
  • anonymous
|dw:1403477725033:dw|
anonymous
  • anonymous
|dw:1403477767271:dw|
myininaya
  • myininaya
if you want to use that specific identity you can replace sin(theta) with 15/17 and solve for cos(theta)
anonymous
  • anonymous
sin(15) = 0.650
anonymous
  • anonymous
sin*17) = -0.96
myininaya
  • myininaya
can you replace sin(theta) with 15/17 in the following equation \[\sin^2(\theta)+\cos^2(\theta)=1\]
anonymous
  • anonymous
no
myininaya
  • myininaya
why not?
anonymous
  • anonymous
well wouldnt you put them in front of sin and cos
myininaya
  • myininaya
just so you know \[\sin^2(\theta)\] and \[(\sin(\theta))^2\] have exactly the same meaning
myininaya
  • myininaya
you are given sin(theta) is 15/17 so you replace the whole sin(theta) thing with 15/17
myininaya
  • myininaya
\[\sin^2(\theta)+\cos^2(\theta)=1 \text{ means exactly that } (\sin(\theta))^2+(\cos(\theta))^2=1\] So we have \[(\sin(\theta))^2+(\cos(\theta))^2=1 \\ (\frac{15}{17})^2+(\cos(\theta))^2 =1 \text{ I did this because I was given that } \sin(\theta)=15/17\]
myininaya
  • myininaya
now solve for cos(theta)
myininaya
  • myininaya
if that confuses you solve the following equation for u \[(\frac{15}{17})^2+u^2=1 \]
myininaya
  • myininaya
whatever value(s) you get for u is your answer since u is cos(theta)
myininaya
  • myininaya
and cos(theta) is the objective
anonymous
  • anonymous
225/285
anonymous
  • anonymous
|dw:1403478332917:dw|
myininaya
  • myininaya
so what did you do to get that?
anonymous
  • anonymous
found cos of theta
myininaya
  • myininaya
you solve the equation (15/17)^2+u^2=1 for u (which is cos(theta) )?
anonymous
  • anonymous
yes
myininaya
  • myininaya
you might want to do that one more time
myininaya
  • myininaya
the denominator is right but your numerator is off
anonymous
  • anonymous
8 / 17
myininaya
  • myininaya
ok so cos could either be 8/17 or -8/17 when sin is 15/17
anonymous
  • anonymous
|dw:1403478613956:dw| my only option
myininaya
  • myininaya
Agreed
anonymous
  • anonymous
In Shadow Bay in July, high tide is at 1:30 pm. The water level is 4 feet at high tide and 0 feet at low tide. Assuming the next high tide is exactly 12 hours later and the height of the water can be modeled by a cosine curve, find an equation for Shadow Bay's water level in July as a function of time (t).
anonymous
  • anonymous
okay so i got f(t) = 4cos pie/2t +1
anonymous
  • anonymous
@myininaya

Looking for something else?

Not the answer you are looking for? Search for more explanations.