Find all points at which two curves intersect. Please help

[Ineedhelp777]

I attached my work. I tried setting them equal to eachother. Then I tried solving sin and cos at one but my answer I got was different fro the one on the key. I will also attach the way it was solved on the key.Its problem #11.

 

[Dr.Peterson]

Please explain what you are doing in your work. It doesn't make sense, as you appear to be just making a guess that doesn't work. (You also appear to be writing "sin(-1)" meaning that the sine of an angle is -1, which suggests you don't fully understand the notation.)

Also, I notice that the problem says "[calculator]", which I take to mean that you are expected to solve it using a graphing calculator or something like it. (That is not necessary, but it appears to be what they have done, since they just give decimal answers. Solving algebraically takes a bit of work.) I don't know why x and y are mentioned.

 

[Ineedhelp777]

If you read the previous comment I explained how to do my work. I am a diligent student, I don't "guess". If you don't understand how to solve the problem that is okay I appreciate your time.

 

[Steven G]

Can you show us your attempt at solving this problem so that we can guide you to the correct answer. If you had read the forum's guidelines you would have know that.

So please post back with your work.

If you can't solve it in polar form then maybe you should try another form.

 

[Ineedhelp777]

I added of attachment of my work of me solving the problem. I indeed read the forum's guideline and already provided the appropriate materials. I don't understand where the miscommunication lies. I added the attachment of my attempt, explained my issue, and added an additional key for reference. Please review my comment again. Thank you.

 

[Steven G]

Your work really did not show much.

HallsofIvy gave you a nice hint which I do not see any more so I will post his brilliant hint.

We can easily obtain that sin(y) + cos(y) = 1/2
We know that sin(x+y) = sin(x)cos(y) + sin(y)cos(x) but we prefer to have sin(y) + cos(y) on the right side or a multiple of it. So let x =pi/4 as sin(pi/4) = cos(pi/4) = sqrt(2)/2

Now we have sin(pi/4 + y) = (sqrt(2)/2)(sin(y) + cos(y)) = (sqrt(2)/2)(1/2) = sqrt(2)/4

That is basically the whole problem. Just solve for y in sin(pi/4 + y) = sqrt(2)/4 using the inverse sin function. Just be careful and get all answers.

 

[Dr.Peterson]

I asked you to explain (which means, in words) what you did.

What I see is that you went from

[MATH]-2\sin\theta - 2\cos\theta = -1[/MATH]​

to

[MATH]-2\sin(-1)[/MATH] and [MATH]-2\cos(-1)[/MATH]​

with no explanation. As far as I can tell, the -1 came from nowhere; but maybe you did something I can't see.

At any rate, when someone doesn't understand what you did, you don't just say "I showed you"; you explain it in a new way. This is how communication works. It may be my fault, or it may be yours, but we have to see more.

I'm wondering if something else you said disappeared from the thread, along with the answer Jomo just referred to.

 

[Dr.Peterson]

Now I see that there are two threads with the same problem; that's part of the problem here. Never do that! It gets very confusing.

But even looking at the other thread, I don't see that you have ever explained the -1.