Basic trigonometric question....

CIV · 2015-10-16 18:27:08

The problem I have to solve is basically an angle in standard position on a Cartesian plane with a terminal end of (4,3). The problem wants me to solve for the six trigonometric functions. I know that I have to solve for the hypotenuse, which is 5. So that means sin=3/5 and so on for the remaining 5 functions. My question is, what about sin=y/r, cos=x/r, and so on? Im confused. So initially I thought I thought sin=3/4, because 3 is or could be the radius on a unit circle and y is... y. Is sin=y/r ONLY for when dealing with a unit circle??? Technically the problem doesn't mention anything about a unit circle. I don't know what is wanted and when or if they are the same thing?!?! I mean clearly 3/5 and 3/4 are not the same thing. I know that sin=opp/hyp and sin=y/r, but it seems to me that those are two different things. What am I missing here? thanks in advance.

Bob · 2015-10-16 20:04:02

hi CIV

The unit circle stuff is explained here:

http://www.mathsisfun.com/geometry/unit-circle.html

Those formulas (sin(angle) = y/r etc.) are true for any sized circle.

The sine of the angle with the x axis is 3/5 so you have got that right.

r stands for the radius, which is 5 in this question.

If you draw a circle radius 5 it will go through that point. Imagine a second circle with a radius of 1. Where does the line from (0,0) to (4,3) cut the smaller circle?

The smaller circle is only one fifth the size of the larger circle but the angle the line makes with the x axis is still the same. Let's call it 'a'. We know sin(a) = 3/5

The line cuts the small circle at (4/5,3/5). So sin(a) = 3/5 over 1 = 3/5. That fraction will never change for this angle 'a'. That's why sines and cosines are so useful. You know the fraction whatever the size of the circle.

If the circle was radius 10. Extend the line to cut the circle at (8,6). Now sin(a) = 6/10. But that simplifies down to 3/5 so the sine(a) is still the same.

Hope that helps,

Bob

CIV · 2015-10-17 03:37:04

Oh my goodness I think I understand this now. So stupid of me. I thought the radius was 3 initially. It's not. 3 is x and 4 is y. I solved the hypotenuse which is the radius. So sin theta = is 3/5. How could I not understand this before! LOL. Like losing your wallet and finding it after realizing it was on the kitchen table right in front of you the whole time. Thanks Bob.

Bob · 2015-10-17 05:11:00

hi CIV

I'm so pleased you have it sorted. Finding wallets / keys / glasses is something I still have to master.

Bob

CIV · 2015-10-17 05:29:02

Thanks bob:) I do have a quick question though. Do you know of a quick way to figure out what quadrant a radian is in without using a calculator?

For example: -17pi / 6

Degree's is easy but radians is a bit challenging without a calculator.

I know I have to find a coterminal angle. I'll add 2pi which will give me -5pi/6, but even with the number looking a bit nicer, I'm still having trouble picturing where this lies within a circle.

I know I most likely need to find a reference angle, but I don't know whether to use pi or 2pi to find it.

I know the answer to this problem, but I had to use my calculator to find out. What is the trick to doing it without a calculator?

Thanks again.

Bob · 2015-10-17 19:27:06

hi CIV,

Working out -5pi/6 is a good place to start. Starting at the positive x axis and rotating 5/6 of pi anticlockwise means you don't quite get to -pi, so this angle must be in the third quadrant. I can go further than this. 1/6 of pi is equivalent to 180/6 = 30 degrees. So this angle is just 30 from the negative x axis.

In general I would add / subtract multiples of 2pi until I had an angle in the range 0 - +/- 2pi and then count round from the positive x axis, 0; pi/2; pi; 3pi/2 etc to see where this angle falls. I usually sketch a diagram too.

Bob

CIV · 2015-10-18 05:15:44

Ok thanks. Ill try this when i get home. I understand a lot of this but cant quite seem to fully wrap my head around it. Having trouble visualizing whats going on. For the most part i understand. Just need to spend more time working with it. Also, is cosine inversely proportional to sine? I noticed the behavior of the two when the triangle moves and it appears they are.

Bob · 2015-10-18 05:38:56

is cosine inversely proportional to sine?

For x and y to be inversely proportional, you need

for some constant k.

It is true that in the range 0 to pi/2, sine gets bigger as cosine gets smaller, but there's no constant of proportionality. If there were then

You will not find a single k to make that happen, even in that range, and outside that range the sign doesn't even work. And things go haywire when sine or cosine goes to zero.

Bob

Math Is Fun Forum

#1 2015-10-16 18:27:08

Basic trigonometric question....

#2 2015-10-16 20:04:02

Re: Basic trigonometric question....

#3 2015-10-17 03:37:04

Re: Basic trigonometric question....

#4 2015-10-17 05:11:00

Re: Basic trigonometric question....

#5 2015-10-17 05:29:02

Re: Basic trigonometric question....

#6 2015-10-17 19:27:06

Re: Basic trigonometric question....

#7 2015-10-18 05:15:44

Re: Basic trigonometric question....

#8 2015-10-18 05:38:56

Re: Basic trigonometric question....

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