alysa111
For an angle Θ with the point (−20, −21) on its terminating side, what is the value of cosine?
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alysa111
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@kelliegirl33 can you help me
myininaya
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If it helps you graph the point. Can you do that for me first?
alysa111
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@myininaya just plot -20,-21 ?
myininaya
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Yep plot the point as a first step.
myininaya
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You can use the draw tool.
alysa111
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ok
alysa111
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|dw:1403474666905:dw|
alysa111
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@myininaya
myininaya
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|dw:1403474755600:dw|
Now I will simply make a right triangle using the center of the graph for my central angle.
myininaya
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what is the ratio definition of cosine of an angle?
alysa111
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legnth of adjacent / legnth of hyp
myininaya
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right you already have the adjacent side given in the picture
the only thing you need now is the hyp of that triangle, right?
alysa111
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yes
myininaya
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Find the hyp of that triangle using Pythagorean Theorem.
alysa111
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the adjacent side is -20
myininaya
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You are correct to say that.
myininaya
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Can you find the length of the hyp?
alysa111
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yes hold on wouldnt it be
-20^2 + 21^2 = c^2
myininaya
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20^2+21^2=c^2
and then take the square root of both sides to give you c.
alysa111
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29
myininaya
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so that is the length of the hyp
so what is cos(theta)=?
alysa111
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29 ?
myininaya
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you said earlier cos(theta) is adj/hyp
you never said cos(theta) is hyp
alysa111
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20/29
myininaya
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don't forget the negative sign
alysa111
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thank you so much i have a few more cn yu help me
myininaya
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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
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|dw:1403475467323:dw|
alysa111
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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
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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
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Well you are talking about shifting the graph up when you have that +5
alysa111
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so the function shifts up 5 units so the range changes from -1 to 1
alysa111
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in f(x) to 4 to 6 in g(x)
myininaya
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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)
alysa111
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am i right
myininaya
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well first what is the range of y=3sin(x)?
alysa111
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f(x) = 3 sin 2x to g(x) = 3 sin 2x + 5 affect the range of the function
myininaya
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yes the range of the graph of f would be affected if you need f+5
myininaya
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did f+5 (not need)
myininaya
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but the range of y=3sin(x) is not -1 to 1.
alysa111
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Which of the following functions has the greatest y-intercept?|dw:1403475803380:dw|
alysa111
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its 2 to 8
myininaya
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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?
alysa111
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yes
myininaya
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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?
alysa111
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so f(x) has greater y-intercerpt
myininaya
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the lowest y value for y=sin(x) is -1
so the lowest y value for f(x)=3sin(x) is ?
alysa111
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-3
myininaya
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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
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now what is the range for y=3sin(x)+5
alysa111
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2 to 8
myininaya
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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
alysa111
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now for Which of the following functions has the greatest y-intercept? wouldnt it be f(x) since the y is 9
myininaya
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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)
alysa111
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Okay , so it would be stated has 2 to 8 in g(x)
myininaya
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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
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I will give you a hint.
All the x values of any point on the y-axis are _____?___
alysa111
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Im not sure
alysa111
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4-9 = -5 which is the same for g(x)
alysa111
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the same y-intercept
myininaya
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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?
alysa111
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on the x axis
myininaya
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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.
alysa111
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so 0-9 ?
0-4 ?
myininaya
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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.
alysa111
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Yeah im confused now
myininaya
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|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?
alysa111
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functions has the greatest y-intercept
alysa111
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1,1 ?
myininaya
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I'm asking you to tell which of the points from f have the x value is 0?
myininaya
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|dw:1403477214481:dw|
alysa111
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its 0,0 ?
myininaya
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|dw:1403477221682:dw|
Do you see any of these x's being 0?
alysa111
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yes
myininaya
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yes the point (0,0) is a y-intercept because the x is 0
myininaya
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now we need to look at g(x)=5cos(3x)-5
myininaya
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what point on g gives us the x is 0?
alysa111
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-5
myininaya
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how did you get that?
myininaya
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Did you replace the x with 0 to see what y is? Because y is not -5 when x is 0.
myininaya
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I only care about when x is 0 since you are asking about the y-intercept.
myininaya
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Replace the x with 0 like so
y-intercept of g is g(0)=5cos(3*0)-5
myininaya
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evaluate 5cos(3*0)-5
alysa111
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5cos(0)-5
5cos
myininaya
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you are right to say 5cos(3*0)-5 can be wrriten as 5cos(0)-5
but where do we go from there?
alysa111
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1.41
myininaya
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:(
5cos(0)-5
what is cos(0)?
alysa111
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1
myininaya
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5(1)-5
can you simplify from here now?
alysa111
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0
alysa111
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same y
myininaya
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right so the y-intercept for g is (0,0) as well
they have the same y-intercept
alysa111
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1f sin Θ = 15 over 17, use the Pythagorean Identity to find cos Θ.
myininaya
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draw a right triangle using this information sin(theta)=15/17
alysa111
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\[\sin^2 \theta + \cos ^2\theta = 1\] will be what we use ?
myininaya
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you can use that
alysa111
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|dw:1403477725033:dw|
alysa111
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|dw:1403477767271:dw|
myininaya
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if you want to use that specific identity you can
replace sin(theta) with 15/17 and solve for cos(theta)
alysa111
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sin(15) = 0.650
alysa111
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sin*17) = -0.96
myininaya
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can you replace sin(theta) with 15/17 in the following equation
\[\sin^2(\theta)+\cos^2(\theta)=1\]
alysa111
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no
myininaya
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why not?
alysa111
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well wouldnt you put them in front of sin and cos
myininaya
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just so you know \[\sin^2(\theta)\]
and \[(\sin(\theta))^2\]
have exactly the same meaning
myininaya
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you are given sin(theta) is 15/17
so you replace the whole sin(theta) thing with 15/17
myininaya
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\[\sin^2(\theta)+\cos^2(\theta)=1 \text{ means exactly that } (\sin(\theta))^2+(\cos(\theta))^2=1\]
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
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now solve for cos(theta)
myininaya
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if that confuses you solve the following equation for u
\[(\frac{15}{17})^2+u^2=1 \]
myininaya
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whatever value(s) you get for u is your answer since u is cos(theta)
myininaya
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and cos(theta) is the objective
alysa111
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225/285
alysa111
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|dw:1403478332917:dw|
myininaya
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so what did you do to get that?
alysa111
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found cos of theta
myininaya
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you solve the equation
(15/17)^2+u^2=1 for u (which is cos(theta) )?
alysa111
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yes
myininaya
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you might want to do that one more time
myininaya
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the denominator is right
but your numerator is off
alysa111
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8 / 17
myininaya
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ok so cos could either be 8/17 or -8/17 when sin is 15/17
alysa111
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|dw:1403478613956:dw| my only option
myininaya
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Agreed
alysa111
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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).
alysa111
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okay so i got
f(t) = 4cos pie/2t +1
alysa111
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@myininaya