ARGH! Okay, So, I Need Smart Math Teachers Advice On This Radian ...

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You are using an out of date browser. It may not display this or other websites correctly.You should upgrade or use an alternative browser. ARGH! Okay, so, I need smart math teachers advice on this radian problem

• Thread starter Peachyness
• Start date Aug 20, 2009

Peachyness

Virtuoso

So, I need to find the exact value for sin (7pi/2) The first thing I always do is convert radian into degrees. Just easier for me. So, 7pi/2 comes out to 630 degrees. So, I know how to find 630 degrees on the circle graph, but I can't seem to figure out if it's a 90-60-30 triangle or if it's a 90-45-45 triangle. So, what am I doing wrong? :thanks: E

EpsilonBeta

Rookie

I would recommend finding out where it lies on the unit circle. If you're in radians subtract 2pi until you're inbetween 0-2pi. If you're in degrees subtract 360 until you're inbetween 0-360. 630 - 360 = 270 OK! We are in between 0 - 360 This is actually not either of your triangles. This is a special point on your unit circle where you get a straight line. You only have sine values and you have no cosine values. 270 is on the bottom of the unit circle therefore we are going down one unit. Or -1. I know degrees is much easier when you're starting out but I would recommend trying to get more comfortable with radians. They end up having nice simple relations to one another based on the fractions. ie any radian over 2 ends up have sine of it be 1 or -1.

Peachyness

Virtuoso

Ah, that makes sense then! Thank you very much. E

EpsilonBeta

Rookie

No problem! I would recommend keeping a copy of a unit circle with you. http://www.math.ucsd.edu/~jarmel/math4c/Unit_Circle_Angles.png Have you used one of these before? x coordinate is the cosine of that angle and the y coordinate is the sine of that angle. It's a great tool to memorize.

Peachyness

Virtuoso

I do have access to one that my calculus teacher posted on her website. But, I still tend to just convert it all to degrees and then go from there. This is the first week and my brain is still a little rusty. But, I will copy that for future use. Thanks! I like this one better than hers anyways. E

ebrillblaiddes

Companion

Instead of using degrees, I would draw a unit circle, then split it into fourths (the entire circle is 2pi, so pi/2 is 1/4 of the circle), and then (this is the hardest part to explain and it'd be so much easier if I could just point, but I'll try) go 7 "steps" around it, counterclockwise as usual, and end up 7/4 of the way around the unit circle, at (0,-1). The sine is the same as the value of y, so sin(7pi/2)=-1 E

ebrillblaiddes

Companion

EpsilonBeta said: This is actually not either of your triangles. This is a special point on your unit circle where you get a straight line. You only have sine values and you have no cosine values. Click to expand...

Actually, the cosine is 0...the unit circle definition of sine and cosine is y and x, respectively. This corresponds with the right triangle definition of sine and cosine for right triangles with hypotenuse 1, for reasons I could explain a lot better with a picture.

Peachyness

Virtuoso

ebrillblaiddes said: Actually, the cosine is 0...the unit circle definition of sine and cosine is y and x, respectively. This corresponds with the right triangle definition of sine and cosine for right triangles with hypotenuse 1, for reasons I could explain a lot better with a picture. Click to expand...

Yeah, I realized what you said before and din't catch that. Now that we are a chapter into calculus, I'm thinking, that problem was a piece of cake!!!! UGH Proofs. BLEH.

Aliceacc

Multitudinous

I LOVE proofs!

Peachyness

Virtuoso

Aliceacc said: I LOVE proofs! Click to expand...

I seem to do well if I have an example in front of me. I guess it's just getting the hang of it. Right now, actually, it's continuities that I need to get a hang of.

Aliceacc

Multitudinous

Anything in particular??

Peachyness

Virtuoso

OH BOY!!! Yes, I have a couple of problems in my calculus book that I am just not getting. Well, since you asked, here is one problem that I didn't understand why I got it wrong. Consider the funtions f(x)= 1, x does not equal 4 .......-1, x=4 and g(x)= 4x-10, x doesn't equal 4 ........-6, x equals 4 In each part, is the given function continuous at x=4? so, I got a, b, c, d, and f, but didn't get e and g. e) f(x)g(x) g) g(x)-6f(x) I wrote that both were not continuous.

Peachyness

Virtuoso

For e, I don't get why it's continuous because, when I chose x to be 4, then this is what I got: x=4, therefore, f(x)=-1, and g(x)=-6 -1*-6 =6, which is not continuous at 4. So.... I must be missing some sort of point here. ??? And then for g) g) g(x)-6f(x) if X=4, then g(x)=-6 and f(x)=-1 So, -6-6(-1) =-6+6 =0 ???? D

dw1

Rookie

It's hard to put this into a text box, but I'll try! Continuous means that if you graph it, you can draw the graph without picking your pencil up. Your original f(x) is not continuous because you are happily drawing your line y=1 for all x except for 4, but when you get to x=4 you have to pick up your pencil and make a dot down at -1, then pick it up again and continue your horizontal line. Now let's look at part e: f(x)g(x) . At every value of x except x=4, f(x)g(x) = 1*(4x-10), which = 4x-10. As this gets closer and closer to x=4 (but not quite there!), f(x)g(x) gets closer and closer to 4(4)-10, which = 6. At exactly x=4, f(x)g(x) = (-1)(-6), which = 6. It's the same number that f(x)g(x) was approaching as it got closer to 4, so it's continuous! If you were drawing the graph with a pencil, your pencil would be getting closer and closer to 6 as x got closer to 4, and right at x=4 your pencil would be in the right spot. You wouldn't have to pick it up and move it. You can do the same analysis for part g: At every value of x except x=4, g(x) - 6f(x) = (4x-10) - 6(-1) which, as x gets closer to 4, approaches g(x) - 6f(x) = 4(4)-10 - 6(-1) = 16 - 10 + 6 = 0. At exactly x=4, g(x) - 6f(x) = (-6) - 6(-1) = -6 + 6 =0. Since the combined function is approaching the same value as x gets closer to 4 as it IS at x=4, it is continuous. I hope this makes sense!

Aliceacc

Multitudinous

It's been a while, so forgive me if I'm rusty on this. In order to be continuous,the following must be true: a) the function must be defined at that point. b) the left and right hand limits at that point must exist, and must equal the value of the function at that point. So, here: f(x)= 1, x does not equal 4 .......-1, x=4 The function exists at 4, it's value is -1 The left and right limits exist. As you get closer and closer to x=4 from the left and right, the value of the function is 4. BUT the two values don't agree. If you were going to graph this, you would have a horizontal line at 1 with a hole at x=4; a that point the value would drop to -1. So the function is discontinous; you wouldn't be able to connect all those y values without stopping, moving your pencis from 1 to -1 and back. Does that make sense??

Peachyness

Virtuoso

oooooooooooooooh okay. I got it. Thanks

mollydoll

Connoisseur

Get the book How to Ace Calculus. It has some pretty good explanations.

Peachyness

Virtuoso

mollydoll said: Get the book How to Ace Calculus. It has some pretty good explanations. Click to expand...

Thanks! I will look into it.

Peachyness

Virtuoso

Is this the book?

mollydoll

Connoisseur

Yes! It is a great bridge between class and the textbook.

Peachyness

Virtuoso

mollydoll said: Yes! It is a great bridge between class and the textbook. Click to expand...

I think I'll get it and see how it is. There were a couple of reviews that said that this is more geared towards high school level calculus. Have you read this book?

mollydoll

Connoisseur

Peachyness said: I think I'll get it and see how it is. There were a couple of reviews that said that this is more geared towards high school level calculus. Have you read this book? Click to expand...

High school and college level calculus are not very different. The material is exactly the same. College courses might add extra topics and cover some proofs and exams may be harder, but calculus is calculus. The examples in the book are solid examples and not easy throw away ones--if you understand the book problems, you are in good shape. For the chain rule, the book shows you how to do problems nested three deep. After you understand that, you can do anything your text is going to toss at you. But, actually, the book is geared mostly towards college, not high school. The concepts are explained well. The thing though, is the book is not going to teach you an entire class. You still need to do lots of practice problems. But the book gives some great techniques and easy explanations to get you started. I used it with all of my (college) tutoring students and it really helped. Which textbook are you using?

mollydoll

Connoisseur

There is also a 2nd book covering multivariable calculus, but it isn't as good.

Peachyness

Virtuoso

Well, I ordered the book. I bought it used, so I figured, why not. the text book I'm using is called Calculus, 9th Edition and it's by Howard Anton, Irl Bivens, and Stephen Davis. I worked on two more sections and got most of the problems figured out.

Peachyness

Virtuoso

So, the book arived today and I must say, that within 3 hours and after reading 9 chapters, I feel a bit better about calculus. I feel that I have that underlying big idea that I've been needing. What's great is that this book teaches in a way that I do. It uses stories and jokes. It tries to make as many connections as it can. That's what makes this book so much different from those textbooks. They don't make it feel personal. Shame, really. I spent $200 on the textbook and $7 total including shipping on this book, and feel that the latter book is way better. Of course, it didn't go over everything the textbook did. But, still, I'm feeling better about calculus (in regards to limits and continuities) Thanks for the suggestion!!! You must log in or register to reply here. Share: Facebook X (Twitter) Reddit Pinterest Tumblr WhatsApp Email Share Link

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