where on unit circle lies sin a =1/3 ?

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Hi, is it possible to determine in which quadrant lies sin a = 1/3 or sin -3/5 without using the calculator to find the angle because in that case , it's very simple. If I don't have access to the calculator , is it possible ?
 
Hi, is it possible to determine in which quadrant lies sin a = 1/3 or sin -3/5 without using the calculator to find the angle because in that case , it's very simple. If I don't have access to the calculator , is it possible ?
If just wanted to know which "quadrant"- then it is possible to guess that, with some additional information (the sign) about the "cosine" or "tangent" of the angle.

If "sine" and "cos" are both positive - then the angle is in first quadrant.

If the sine is negative and cosine is positive, then the angle is in fourth quadrant. These come from the "unit circle".

To get the "value" of the angle, you have to get a "sine" table where a large number sine values are listed ( and interpolate).
 
Hi, is it possible to determine in which quadrant lies sin a = 1/3 or sin -3/5 without using the calculator to find the angle because in that case , it's very simple. If I don't have access to the calculator , is it possible ?
You have things backward.

The calculator can't tell you what quadrant angle a is in! That's something you need to do by thinking ... because there are two possible quadrants in each case, so you need to either list both, or use additional information.

If you ask the calculator for the answer (that is, the value of a, whose quadrant you can determine), it will give you only one answer, the inverse sine, because that's all it can do. (That's always in the first or fourth quadrant.) Even with the calculator, you need to do the thinking to apply its answer.
 
You have things backward.

The calculator can't tell you what quadrant angle a is in! That's something you need to do by thinking ... because there are two possible quadrants in each case, so you need to either list both, or use additional information.

If you ask the calculator for the answer (that is, the value of a, whose quadrant you can determine), it will give you only one answer, the inverse sine, because that's all it can do. (That's always in the first or fourth quadrant.) Even with the calculator, you need to do the thinking to apply its answer.
"The calculator can't tell you what quadrant angle a is in! "- No, i meant to say that if i use the calculator to evaluate for sin a =1/3 by doing inverse sine 1/3 and i get the angle , let say 20, then i know it's in quadrant 1.
 
"The calculator can't tell you what quadrant angle a is in! "- No, i meant to say that if i use the calculator to evaluate for sin a =1/3 by doing inverse sine 1/3 and i get the angle , let say 20, then i know it's in quadrant 1.
Incomplete answer - it could also be (180-20 = ) 160o or (360 + 20 = )\(\displaystyle \cancel {160}\)o 380o .. corrected..... and so on......
 
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No, i meant to say that if i use the calculator to evaluate for sin a =1/3 by doing inverse sine 1/3 and i get the angle , let say 20, then i know it's in quadrant 1.
The point I made was precisely that this isn't the only answer. It can tell you a possible quadrant, but not the quadrant (which can't be known with only this information).

The mere fact that the sine is positive, all by itself, tells you that the angle must be in quadrant 1 or 2; and if it were negative, the angle would be in quadrant 3 or 4. The calculator would only tell you 1 or 4, respectively.

That should be 380. And we can find more by adding 360 to 160 too right?
Adding more 360's won't change the quadrant, so that's not really part of the answer to your question, though you're right. The important part is 180 - 20 = 160, which gives you quadrant 2 as well as quadrant 1. Do you understand that part?
 
The point I made was precisely that this isn't the only answer. It can tell you a possible quadrant, but not the quadrant (which can't be known with only this information).

The mere fact that the sine is positive, all by itself, tells you that the angle must be in quadrant 1 or 2; and if it were negative, the angle would be in quadrant 3 or 4. The calculator would only tell you 1 or 4, respectively.


Adding more 360's won't change the quadrant, so that's not really part of the answer to your question, though you're right. The important part is 180 - 20 = 160, which gives you quadrant 2 as well as quadrant 1. Do you understand that part?
Yup i got it thanks.
 
If I am given something instruction like this : Determine graphically the arc α such that sin α = 1/3 with 0 < α < pi/2 ?
 
If I am given something instruction like this : Determine graphically the arc α such that sin α = 1/3 with 0 < α < pi/2 ?
You can't determine an exact value graphically, and probably wouldn't be expected to find a close approximation, but you could draw the figure as accurately as you can and estimate the angle:

1656597120036.png

As you can see, there are two angles (BAD and BAC) with this sine; these are the two points on the unit circle with y=1/3. The quadrant tells you to choose 3. But if you expected a question like this on a test, you should ask the instructor what sort of answer would be expected.
 
You can't determine an exact value graphically, and probably wouldn't be expected to find a close approximation, but you could draw the figure as accurately as you can and estimate the angle:

View attachment 33266

As you can see, there are two angles (BAD and BAC) with this sine; these are the two points on the unit circle with y=1/3. The quadrant tells you to choose 3. But if you expected a question like this on a test, you should ask the instructor what sort of answer would be expected.
No they are not asking for the exact value or even the estimation of the angle, just the arc that is made on the unit circle. In the next exercise after this one, they want to us to use the unit circle and this method to find how many solutions are there in the given interval . SO I presume they want us to able to determine where a trig expression is on the unit circle to find the number of solutions. I did three questions following what you did and I am attaching it here. For the first question I did sin a = -(sqrt)(2) /2 the interval given is -pi/2 < a < 0 . But the arc is not in the given domain. Does that mean no solutions?
 

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No they are not asking for the exact value or even the estimation of the angle, just the arc that is made on the unit circle.
The length of an arc on the unit circle is the radian measure of the angle! The question is, what sort of answer are they asking for? Did you quote an actual problem as given to you?

In the next exercise after this one, they want to us to use the unit circle and this method to find how many solutions are there in the given interval . SO I presume they want us to able to determine where a trig expression is on the unit circle to find the number of solutions. I did three questions following what you did and I am attaching it here.
My diagram was actually intended to supplement what had been said previously about how many solutions there are, in addition to suggesting what you could do for the problem you stated (which I took to be one you made up with an unclear statement). So I was really helping you more with this new problem than with the one you asked about!

For the first question I did sin a = -(sqrt)(2) /2 the interval given is -pi/2 < a < 0 . But the arc is not in the given domain. Does that mean no solutions?
For this first problem, there is an arc in the given domain:

1656604277659.png

You show two possible terminal rays, each of which defines two possible arcs in standard position (clockwise and counterclockwise). You need to look for one that is in the right domain, not just pick one and see if it is. I'm presuming you had this arc in mind:

1656604454130.png

You don't show your actual answer for any of these, so I can't be sure if you are thinking correctly.

Also, given that graphs are included in the problems, I'm wondering if when they ask "Determine graphically the arc", they mean just to show the arc on the graph, not to give a numerical answer. (This is part of the reason we like to see the entire problem!) To do that, you'd need to include what I drew in above, for the one requested arc. What you drew only shows that in each case there are two solutions if there were no restriction.
 
The length of an arc on the unit circle is the radian measure of the angle! The question is, what sort of answer are they asking for? Did you quote an actual problem as given to you?


My diagram was actually intended to supplement what had been said previously about how many solutions there are, in addition to suggesting what you could do for the problem you stated (which I took to be one you made up with an unclear statement). So I was really helping you more with this new problem than with the one you asked about!


For this first problem, there is an arc in the given domain:


You show two possible terminal rays, each of which defines two possible arcs in standard position (clockwise and counterclockwise). You need to look for one that is in the right domain, not just pick one and see if it is. I'm presuming you had this arc in mind:


You don't show your actual answer for any of these, so I can't be sure if you are thinking correctly.

Also, given that graphs are included in the problems, I'm wondering if when they ask "Determine graphically the arc", they mean just to show the arc on the graph, not to give a numerical answer. (This is part of the reason we like to see the entire problem!) To do that, you'd need to include what I drew in above, for the one requested arc. What you drew only shows that in each case there are two solutions if there were no restriction.
I am attaching the question down below. They are just asking for the arc a. I drew the arcs like you said, but I am finding it difficult to interpret the interval notations. I know that negative angles mean I have to turn clockwise direction and for positive angles I have to turn anticlockwise.
 

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I am attaching the question down below. They are just asking for the arc a. I drew the arcs like you said, but I am finding it difficult to interpret the interval notations. I know that negative angles mean I have to turn clockwise direction and for positive angles I have to turn anticlockwise.
Here is how I would do the third one:

1656609695727.png

Note that I labeled negative angles, starting clockwise from zero, namely -pi/2, then -pi. This is like negative numbers on a number line; they increase as you go counterclockwise. Having them labeled makes it easier to see that the angle in quadrant 3 is in the interval you want. The other one is between -pi/2 and 0.
 
Hi, is it possible to determine in which quadrant lies sin a = 1/3 or sin -3/5 without using the calculator to find the angle because in that case , it's very simple. If I don't have access to the calculator , is it possible ?
sin a = opposite/hypotenuse, since the hypotenuse is always positive the sign of sin a is determined by the sign of the opposite side. That is a y-value. That is sin a = y and sin a = 1/3. Then y= 1/3 >0 which only happens in quadrant I and II.
 
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