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@kelliegirl33 can you help me

If it helps you graph the point. Can you do that for me first?

@myininaya just plot -20,-21 ?

Yep plot the point as a first step.

You can use the draw tool.

ok

|dw:1403474666905:dw|

what is the ratio definition of cosine of an angle?

legnth of adjacent / legnth of hyp

yes

Find the hyp of that triangle using Pythagorean Theorem.

the adjacent side is -20

You are correct to say that.

Can you find the length of the hyp?

yes hold on wouldnt it be
-20^2 + 21^2 = c^2

20^2+21^2=c^2
and then take the square root of both sides to give you c.

29

so that is the length of the hyp
so what is cos(theta)=?

29 ?

you said earlier cos(theta) is adj/hyp
you never said cos(theta) is hyp

20/29

don't forget the negative sign

thank you so much i have a few more cn yu help me

|dw:1403475467323:dw|

Well you are talking about shifting the graph up when you have that +5

so the function shifts up 5 units so the range changes from -1 to 1

in f(x) to 4 to 6 in g(x)

am i right

well first what is the range of y=3sin(x)?

f(x) = 3 sin 2x to g(x) = 3 sin 2x + 5 affect the range of the function

yes the range of the graph of f would be affected if you need f+5

did f+5 (not need)

but the range of y=3sin(x) is not -1 to 1.

Which of the following functions has the greatest y-intercept?|dw:1403475803380:dw|

its 2 to 8

yes

so f(x) has greater y-intercerpt

the lowest y value for y=sin(x) is -1
so the lowest y value for f(x)=3sin(x) is ?

-3

now what is the range for y=3sin(x)+5

2 to 8

Okay , so it would be stated has 2 to 8 in g(x)

I will give you a hint.
All the x values of any point on the y-axis are _____?___

Im not sure

4-9 = -5 which is the same for g(x)

the same y-intercept

on the x axis

so 0-9 ?
0-4 ?

Yeah im confused now

functions has the greatest y-intercept

1,1 ?

I'm asking you to tell which of the points from f have the x value is 0?

|dw:1403477214481:dw|

its 0,0 ?

|dw:1403477221682:dw|
Do you see any of these x's being 0?

yes

yes the point (0,0) is a y-intercept because the x is 0

now we need to look at g(x)=5cos(3x)-5

what point on g gives us the x is 0?

-5

how did you get that?

Did you replace the x with 0 to see what y is? Because y is not -5 when x is 0.

I only care about when x is 0 since you are asking about the y-intercept.

Replace the x with 0 like so
y-intercept of g is g(0)=5cos(3*0)-5

evaluate 5cos(3*0)-5

5cos(0)-5
5cos

you are right to say 5cos(3*0)-5 can be wrriten as 5cos(0)-5
but where do we go from there?

1.41

:(
5cos(0)-5
what is cos(0)?

5(1)-5
can you simplify from here now?

same y

right so the y-intercept for g is (0,0) as well
they have the same y-intercept

1f sin Θ = 15 over 17, use the Pythagorean Identity to find cos Θ.

draw a right triangle using this information sin(theta)=15/17

\[\sin^2 \theta + \cos ^2\theta = 1\] will be what we use ?

you can use that

|dw:1403477725033:dw|

|dw:1403477767271:dw|

sin(15) = 0.650

sin*17) = -0.96

can you replace sin(theta) with 15/17 in the following equation
\[\sin^2(\theta)+\cos^2(\theta)=1\]

no

why not?

well wouldnt you put them in front of sin and cos

just so you know \[\sin^2(\theta)\]
and \[(\sin(\theta))^2\]
have exactly the same meaning

you are given sin(theta) is 15/17
so you replace the whole sin(theta) thing with 15/17

now solve for cos(theta)

if that confuses you solve the following equation for u
\[(\frac{15}{17})^2+u^2=1 \]

whatever value(s) you get for u is your answer since u is cos(theta)

and cos(theta) is the objective

225/285

|dw:1403478332917:dw|

so what did you do to get that?

found cos of theta

you solve the equation
(15/17)^2+u^2=1 for u (which is cos(theta) )?

yes

you might want to do that one more time

the denominator is right
but your numerator is off

8 / 17

ok so cos could either be 8/17 or -8/17 when sin is 15/17

|dw:1403478613956:dw| my only option

Agreed

okay so i got
f(t) = 4cos pie/2t +1