H Alg 2 Trig Ch. 13 - Trigonometry - Dr. Johnson's Information Station

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H Alg 2 Trig Ch. 13 - Trigonometry

2/28/2011

Happy Blogging!

Archana Sathappan

3/1/2011 10:48:25 am

For number 11 pg 717, I got 46 degrees for the reference angle and 134 degrees for the angle in standard position. So when I graph it, should my arrow go in the negative or positive direction...since the original given angle is -2746 degrees.

Archana Sathappan

3/1/2011 11:39:48 am

For numbers 19 and 20, which ways are the arrows supposed to go?

Randy Boyd

3/3/2011 11:52:57 am

For # 18 on page 726, when you make the coordinates for when the angle is 270, (0,-1) is it read as (cosine,sine) or (sine,cosine)?

Ishta

3/6/2011 02:00:05 am

Archana, I think the reference should be drwn in the positive direction because the numbers are positive, I think. Check with other people as well. That is what I put.

Archana Sathappan

3/6/2011 04:42:20 am

I am a bit confused on p 726 number 5...can someone plz help me out?

Brandon Jachimiec

3/8/2011 07:24:04 am

Archana, For page 726 #5,
You use the cordinate point (0,-1) which is also (cosine, sine) so:

the sin of 180 is 0 (from cordinate pt)
the cosine of 180 is -1 (from cord. pt)
the tangent is 0 (0/-1)
the cosecant is undefined (-1/0)
The secant is -1 (1/-1)
The cotangent is undefined (1/0)

Ishta

3/8/2011 08:32:10 am

On the 13-5 homework, does the theta sign mean the same thing as x, so like y=cos(theta) is the same as y=cos(x)?

Archana Sathappan

3/9/2011 08:58:13 am

Ishta,
I'm pretty sure you graph them the same way.

Christian Carvallo

3/9/2011 09:31:30 am

I'm so lost can some one explain to me how you get the period in these trigonometry graphs.

Anna Reimers

3/9/2011 09:35:49 am

on pg. 749 #5 i understand you move the graph up 3 and the amp is 2. But what about the 1/5? Should you multiply that by (x-pi) ?

Archana Sathappan

3/9/2011 09:48:54 am

Christian,
Periods for sine and cosine are found by 2pi/absolute value of b. For tangent and cotangent you find it by pi/absolute value of b.

Archana Sathappan

3/9/2011 09:50:07 am

I need help to graph number 5 pg.749. Can someone plz help me?

Kristina Chapman

3/9/2011 11:18:13 am

Anna, yeah i just had that be the "b" value when figuring out the period for the graph. and if anyone knows how to graph number five, i'm so confused on it!

Melina Kanji

3/9/2011 01:08:59 pm

So for tonight's homework, on page 749, when we are trying to figure out the period, and for example, if we're doing number 5, it says y= 3+ cos1/5 (x-π), would the coefficient of x, be 1/5? or would it just be 1? so to figure out the period, would you distribute the 1/5, then solve?

Archana Sathappan

3/9/2011 09:04:19 pm

Melina,
To find the period for that particular problem you would have to do 2pi/absolute value of since it's cosine. So you would do 2pi/(1/5)...which then you would do freeze and flip, so you get 2pi*5...period is equal to 10pi.

Noelle

3/10/2011 07:27:36 am

I was wondering, are the asymtotes only for secant, cosecant, and cotangent?

Archana Sathappan

3/10/2011 08:03:18 am

Noelle,
There are asymptotes for all of the ones that you mentioned AS WELL AS tangent.

Tyler Yanisch

3/10/2011 02:51:05 pm

Can i get some help on #8 on page 749

Noelle Linden

3/11/2011 02:46:33 am

for writing the equations, how do you find the horizontal shift?

Archana Sathappan

3/12/2011 09:26:32 am

Can someone please tell me what the equation is for number 4 pg 752 and show me how to do a couple of the steps? Because for the answer, I got y=-10cos pi/10(x+7)-40. But, when I put it in the calculator to check it, it's not coming out right.

Archana Sathappan

3/12/2011 09:32:14 am

Sorry! I meant to say pi/15, not pi/10.

Christian Carvallo

3/13/2011 05:39:52 am

On skills practice 105 worksheet how do you graph the equations where it's phase shift is in degrees, yet it's period is in radians. Thanks

Sydney Stevens

3/13/2011 10:31:27 am

Noelle you find the horizontal shift by using the number thats inside the parenthesis (the opposite of it). and I had the same question as Christian too so if anyone can help with that thankyou!

Erin Hohman

3/13/2011 02:26:24 pm

For the worksheet skills practice 105 number 9...how do you graph a period of 1?

Archana Sathappan

3/13/2011 02:53:26 pm

For skills practice 106 number 2, I am confused on which points to choose in order to make an equation.

Emma Burck

3/14/2011 08:56:00 am

Archana: You could really start on any of the points and be able to write the equation, but what made most sense to me was starting at -30 and ending at 150 so that you have a negative sine graph.

Erin: To graph a period of 1 just have your starting point as 0 and the end point of 1 and .5 will be the halfway point. Its basically the same as a graph with pi or degrees except its just simple numbers, there's nothing different you really have to do.

Hope this helps :)

Sam bostick

3/14/2011 09:16:28 am

to Christian: to find the period in radians you use the equation (2π/b). and 2π=360 degrees so the equation to find the period in degrees is (360/b).

Zain Rahman

3/14/2011 09:38:52 am

christian: i think you can change the period to degrees also and just graph it that way. Like in the skills practice 105 for number 1 the period was pi, (which equals 180 degrees). its going to be the same thing either way then you just move the numbers around to do the horizontal shift.

Ishta

3/14/2011 11:04:27 am

For the math models, I still can't get how to plug an equation in? Can someone help me?

Ishta

3/14/2011 11:05:00 am

In the calculator that is.

Amir Khalid Richardson

3/14/2011 11:37:23 am

@ Ishta
to graph in the calculator, you would have to plug the equation in the general form: y= A(cos/sin)(Bx-BC)+D
you have to distribute the B-value to the parenthesis and have the right amount opened and closed

Nikita Hariharan

3/14/2011 04:54:32 pm

I am confused on #2 on Pg. 760 especially b and d. How do u solve this problem?

Ishta

3/15/2011 10:22:50 am

For the inverse circular functions homework, I do not understand how the arc sin, arc cos, arc tan and so forth. Can someone explain it to me?

Archana Sathappan

3/15/2011 01:16:51 pm

Ishta,
They are all on page 770 in the textbook. Sine and tangent, you have the 90 degree arc thing in the first and fourth quadrant and for cosine you have the 180 degreen arc thing that covers the first and second quadrant.

Nidhi Chintalapani

3/15/2011 02:38:23 pm

I don't understand the last 3 problems in the worksheet for 13-7. I got the equations for the rest of them though but i just can't get the last ones.

Rohan Roy

3/15/2011 02:41:44 pm

To Ishta,

Basically like Archana said, you have to see what the question is asking for, like if they are asking for the arcsine root2/2 the answer is Pi over 4, and it is only Pi over 4 since it can only be in the first an fourth quadrant.

Anna Reimers

3/15/2011 03:19:29 pm

Nikita - to find the equation break it down into the different parts. To find the period you find the distance between 0.3 seconds and 1.8 seconds because that is one cycle. Then you can find the other parts from the graph since you know the highest point is at 60 and the lowest is at 40. Then for part d. you move the graph over so it starts at 0 instead of 0.3 in order to find the distance.

Brandon Jachimiec

3/16/2011 08:54:47 am

On page 779 for the questions like #70, or 71, how do you choose between degrees or radians? Is it radians because their is no degree sign, or is there something else to it? I know, I can just do both then see which one is write in the back of the book, but I was just wondering if I was forgetting something.

Nidhi Chintalapani

3/16/2011 12:29:00 pm

Brandon- I'm pretty sure you use degrees because in class that is what we've been using for this section. And Mrs.Johnson has been using theda, so i'm pretty sure it's degrees. But I'm not entirely sure.

Craig Cook

3/16/2011 12:59:17 pm

To Nidhi,
Pertaining to #7 on the 13-7 worksheet. Y= 3cosπ/24(x-6)-1. the 3 refers to the amplitude. I use cosine from (6,2) to (42,2). The period is 48 and half 48 is 24 so it is π/24. I put in a -6 because the graph moves 6 units to the right. The vertical shift is down 1 because the ‘center’ of the graph moves down 1 from y = 0. by center I mean if you drew a dotted line horizontally through the whole graph. In this case the height of the graph from 2 to -4 is 6. the middle between the 2 points is at -1. dotted line at y = -1. I hoped that helped if you haven’t figured this out yet.

Kristina Chapman

3/16/2011 01:14:34 pm

Brandon and Nidhi-
Yeah, I used degrees for number 71, checked in the back of the book, and it worked. Plus, like Nidhi said, Mrs. Johnson has been using degrees in the equations during class. And on number 95 for page 779 I was wondering how you would plug it into the calculator in order to check it. For this problem, would it be 1/(tan^-1(tan 79)? because when I try to plug this into the calculator, I get a different answer than that of the book.

Archana Sathappan

3/16/2011 01:19:39 pm

I skipped problem 2 on pg. 761 b/c I didn't get it, and I still don't get it...like, how to graph it...I'm kinda confused on the graphing part, but I think that I got the rest with setting up the equation and all. The graph is the part that I need help with.

Rohan Roy

3/16/2011 02:45:57 pm

For problem number 81 on page 779, I dont understand how they are getting -1 as an answer. Doesnt the cotangent mean it is cos/sin?

Anna Reimers

3/16/2011 03:26:14 pm

Rohan - The cotangant also means 1/tangent correct? The angle is 315 degrees, so the tangent is -1, so the cotangent would be 1/-1 which equals -1.

ryan v

3/17/2011 01:00:17 pm

anyone know how to do number 69 on pg 779, or any problem similar to that?

Randy Boyd

3/17/2011 01:26:09 pm

Ryan, you just plug Sine(-0.9692) into the calculator. i believe the back of the book gives the answer in radians though.

Randy Boyd

3/17/2011 01:58:31 pm

How do you put the equations of the math models in the calculator. my graph never shows up

Andy

3/17/2011 02:01:08 pm

Randy you have to distribute the period so your calculator knows through what 'time' per say to graph a full cycle through, if it's still not showing up zoom in or out

Randy Boyd

3/17/2011 02:03:20 pm

ooo, gotcha. Thanks Andy

sahana s

3/17/2011 03:13:36 pm

so when trying to find the equation of a graph the amplitude is from the axis to the high node but to get the real axis (vertical shift) its just the middle of the top and bottom most coordinate points?

Nikita Hariharan

3/17/2011 03:16:29 pm

You subtract or add to the amplitude so the placement would be different on the graph.

Nikita Hariharan

3/17/2011 03:17:02 pm

Sahana: Yes.

Nikita Hariharan

3/17/2011 03:18:52 pm

Is the answer to R3) d. 1/pi?

Sahana S

3/17/2011 03:21:55 pm

no nikita because the cosine of 180 degrees is -1 and secant is 1 over cosine so 1 over -1 = -1

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H Alg 2 Trig Ch. 13 - Trigonometry

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