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#1 Re: Help Me ! » Expressing Trig functions in terms of other Trig functions » 2021-06-18 05:52:35

Hi Bob,

I think I understand! Here is what I did:


1/cot = sin/cos

sin²θ + cos²θ = 1     (a trig identity I learned in class)
-> cos²θ = 1 - sin²θ
-> cosθ = √(1 - sin²θ)

So therefore

sin/cos = √(1 - sin²θ)
So,
1/cot = √(1 - sin²θ)

Did I do this right?


I also have another question that I am confused about.

14. Find the principal value of Arctan (-1.44) to the nearest minute.

I have NO clue where to start on this, I don't even know what a principal value is. Could you maybe show me an example of how to solve a question similar to this so that I see the steps of how to find a principal value?

Thank you!

#2 Help Me ! » Expressing Trig functions in terms of other Trig functions » 2021-06-17 05:38:44

camicat
Replies: 5

Hello all,

I am struggling with this question:

7. What is 1/cot(x) in terms of sine?

Here is what I have so far, please tell me if I'm starting off incorrectly!

1/cot(x) = tan
tan = sin/cos
1/cot = sin/cos


And then I'm not sure where to go from here. I am supposed to be using the basic identities listed below to help me express 1/cot in terms of sine:

cscΘ = 1/sinΘ
secΘ = 1/cosΘ
cotΘ = 1/tanΘ



sinΘ = 1/cscΘ
cosΘ = 1/secΘ
tanΘ = 1/cotΘ


sinΘ * cscΘ = 1
cosΘ * secΘ = 1
tanΘ * cotΘ = 1
Ratio Identities



sine = opposite / hypotenuse
cosine = adjacent / hypotenuse
tangent = opposite / adjacent


Here is an example of one problem solved correctly in the way my teacher wants me to do it:

11. Express tangent in terms of cosine.

secΘ = 1/cosΘ

->Tan^2Θ + 1 = sec^2Θ

->Tan^2Θ = sec^2Θ – 1

->Tan^2Θ = (secΘ +1) (secΘ -1)

-> Tan^2Θ = (1/cosΘ +1) (1/cosΘ -1)

-> Tan^2Θ = ([1/cosΘ] +[cos Θ /cos Θ]) ([1/cosΘ] –[cos Θ/cos Θ])

-> Tan^2Θ = (1+cosΘ /cosΘ) (1 -cosΘ /cosΘ)

-> Tan^2Θ = (1-cos^2Θ /cosΘ)

-> √Tan^2Θ = √(1-cos^2Θ /cosΘ)

-> tanΘ = √(1-cos^2Θ) /cosΘ

Tangent in terms of cosine is tanΘ = √(1-cos^2Θ) /cosΘ


I hope that someone could give me some direction on where to go from here. We did this earlier in the year and now I am having to solve this problem as part of a review and I just can't remember how to do it. Thanks everyone smile

#3 Help Me ! » Inverse Trig Functions » 2021-05-10 06:07:12

camicat
Replies: 8

Hello,
I am working on Inverse Trig Functions in school and I am honestly super frustrated with it. My teacher keeps giving me videos to watch, but it's really hard for me to focus on videos. I would greatly appreciate it if someone could write out the steps on how to solve these problems so that I can understand the process of it. Even just walking through a practice example for each type of question would be super helpful. Here are some examples of the different types of questions I am working on:


2. Find the principal value of each of the following in radians:

Arccsc (-2)

A. pi/2
B. -pi/6
C. pi/5
D. -pi
E. 5pi/6
F. -2pi


6. Find the principal value of each of the following to the nearest minute:

Arccsc (1.607)

A. 38°64'
B. 39°24'
C. 38°48'
D. 38°32'
E. 39°37'
F. 38°29'


11. Solve for the angle x.

2cos(x) - 2 = -1

A. no solution
B. pi/3 + 2pin
C. pi/3 + 2pin, 5pi/3 + 2pin
D. 5pi/3 + 2pin
E. pi/3
F. pi/3, 5pi/3

Thanks in advance.

#4 Re: Help Me ! » Rewriting Trigonometric Functions » 2021-04-02 05:46:57

Hi,
That helped a lot! Thank you! I have one last thing that I am stuck on. Here is the problem:

20. Verify (sinΘ)^4 + 2(sinΘ)^2(cosΘ)^2 + (cosΘ)^4 = tanΘcotΘ

I have factored it down so that (sinΘ)4 + 2(sinΘ)2(cosΘ)2 + (cosΘ)4 = (sinΘ^2+cosΘ^2)^2, I'm just not sure where to go from here.

#5 Re: Help Me ! » Rewriting Trigonometric Functions » 2021-03-31 04:59:23

Hi Bob,
I'm still very confused. This is how my teacher wants me to look at it (and how my answer should be formatted):


write cosine in terms of tangent, in a way that avoids using sine. One of the other ways to write cosΘ is to say 1/secΘ

We know that tan2Θ + 1 = sec2Θ

This means that secΘ = √(tan2Θ + 1)

So cosΘ = 1/√(tan2Θ + 1)

Putting this together with what we got before we wind up with an answer:

sinΘ = tanΘ * 1/√(tan2Θ + 1)


Can you help me understand how to figure out my answer in that format?

#6 Help Me ! » Rewriting Trigonometric Functions » 2021-03-30 02:04:43

camicat
Replies: 9

Hello, it's been a long time since I posted last! I am very stuck on this, and could use some help.

I have been tasked with this question:

12. Express cosecant in terms of tangent.

I think I might have a start, but I'm not sure if I'm correct. Here is what I have so far:

tan = sin/cos
sin= 1/csc
tan= (1/csc)/cos

That's as far as I can think. I also have some other, similar problems that I'm struggling with.

13. Express sine in terms of cotangent.


14. Express cosecant in terms of cosine.


15. Express secant in terms of sine.

Any help would be greatly appreciated smile

#7 Help Me ! » Functions » 2020-05-20 04:08:29

camicat
Replies: 1

Hello all,

I am having incredible difficulty understanding these problems. I understand the definition of domain and range, however I do not understand how I am supposed to find them from these given problems. On top of that, I have no idea how I am supposed to know how to draw the relation on a graph. Any help would be greatly appreciated!


C.  State the domain and range: Use the grid to draw the graph and verify your answer.  Be sure to label the scales on your axes to accommodate the numbers in the problem. (You may not use a computer program to draw the graphs for you.  The goal is to understand how to select points to draw a graph.)

5.  y = x^2 + 86

6.  y = 2x^3 + 8

7.  y = 40 - 8x^2


I see how they are in slope-intercept form, however I do not know how to graph them, because they all have exponents attached to x, and I don't know how that affects the graph.

Thanks in advance!

#8 Re: Help Me ! » Radicals and Roots » 2020-04-06 07:43:28

Hi, Bob!

This was very helpful, thank you!

#9 Help Me ! » Radicals and Roots » 2020-04-03 06:39:24

camicat
Replies: 2

Hello, I'm once again posting because I need help with a word problem. I can't even decipher what I need to solve. Here is the word problem, thanks in advance for any help!


10. A weight hangs on a spring. The vibration period of that weight is the time in which it makes a complete cycle in motion. Suppose the relationship between the vibration period "T" (in seconds) and the weight "w" (in kilograms) is given by T = 2 pi SQRT(w / 200). Find the period T, to the nearest thousandth, for a spring with a hanging weight of 2.0 kilograms.

#10 Re: Help Me ! » Algebra 2 Factoring Polynomials » 2020-03-19 05:05:03

Hello Bob,

Thank you for your reply! I am still a bit confused, though. It has been a while since I have done basic algebra, so I feel that I may just be forgetting how to do this. Here is what I have done so far:

h = -3.9t^2 + 15.6t
0= -3.9t^2 + 15.6t
0 = 3.9 (-t^2 + 4t)
0 = -3.9t (t + 4)

I feel like I have not found the value that I am supposed to find, but I honestly don't even know what value I am trying to find. I am honestly just very confused and frustrated with myself for feeling like I have forgotten how to do a simple problem. I would appreciate any help you could give me with straightening out what I am doing.

#11 Help Me ! » Algebra 2 Factoring Polynomials » 2020-02-27 08:21:15

camicat
Replies: 3

Hello,
I (mostly) understand how to factor polynomials, but word questions always trip me up! Here is the current question that is confusing me:


Suppose a football is kicked from the ground and its height, h, in feet above the ground is given by h = -3.9t2 + 15.6t.

The time, t, represents the number of seconds after the ball is kicked. At what time does the football hit the ground?


Thanks in advance smile

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