# Help with some trigonometry problems please

1. Mar 13, 2004

### mathzeroh

help with some trigonometry problems please!!

hello everybody! how's everyone doing? i hope you guys are well and your homework is going great!
i posted another help thread but it was in the wrong section! sorry! but i got great help there!

ok here's my new problem. i tried, but i just don't know how i could do it! it's a bit long but please bear with me.

Write each rectangular equation in polar form.
26. y=-5
27. x=10
28. x^2+y^2=7 (latex version: $$x^{2}+y^2=7$$)
29. 2x^2+2y^2=5y (latex version:$$2x^{2}+2y^{2}=5y$$)

This is what i did:
26: y=-5 ===> I used the pythagorean theorem: r=&radic;(x^2+y^2)
r=&radic;(x^2+(-5)^2) ---> I plugged in the value for y into this formula
r=&radic;(x^2+25) ---> -5 squared is 25
r=5+&radic;(x^2) ---> 25 square rooted is 5
r=5+x ---> the square root of x squared is x...isn't it?[?]

and also, i said that &theta;=Undefined because they did not give me an "x" value, so i just said that x=0; because, the formula

for converting a rectangular coordinate to polar coordinate is:
$$\theta=tan^{-1}(y/x)$$, when x>0

and x cannot be zero in the denominator spot right?

and that's what my answer is. but i have this feeling that it is not right! because i did the next problem which is very similar

to this one. (and sadly the answers are not in the back of the book for even number problems)

27: basically i did the same exact thing as number 26, and got this:
r=10+y --->again, here, &theta;=zero because of the same reason up there (except this time i didn't have a "y" value, so i said

y=0 and got zero because arctan 0=0).

so i checked in the back to see if this one was right, but it wasn't.[b(] [b(] this is what was there (the answer for #27)
r=10sec&theta;

HELP!!

then, i got to number 28, and i didn't have a clue as to how to even begin this problem! same with 29! can someone please

help me with these? any help is appreciated! thanks a lot in advance!

Last edited: Mar 13, 2004
2. Mar 13, 2004

Whew! I think you're working way too hard!

Okay, let's start from the beginning. When we have some coordinate pair (x,y), we draw a right triangle by dropping a line straight down (or up, if y is negative) to the x-axis. Then we have a nice little triangle.

Now let's figure out what we know from that triangle. Let's say we were given r and theta and we wanted to find x and y. How would we do it? Well, we know that sin = opposite/hypotenuse and cos = adjacent/hypotenuse, so let's see where that takes us.

$$\sin{\theta} = \frac{\textrm{opposite}}{\textrm{hypotenuse}} = \frac{y}{r}$$

which we can rearrange to get

$$y = r\sin{\theta}$$

You can do the same with the cosine to get an expression for x.

So how does this help you? Well, let's say that y = -5. So what happens when we plug that into the equation we just got?

$$y = r\sin{\theta}$$
$$-5 = r\sin{\theta}$$
$$r = \frac{-5}{\sin{\theta}}$$

That's what we want, right? A function for r in terms of theta. All done!

Now try it for 27, except this time use the equation you derived for x using cosine.

As for 28 and 29, I think you should try the method you tried to use for 26 and 27 in your post. I think I see some x^2 + y^2's in there that will turn into some nice r^2.

So the moral of the post: if you see a lonely y or a lonely x, use r*sin(theta) or r*cos(theta) (respectively), but if you see x^2 + y^2's, turn them into r^2's!

Why don't you give them all a try again and see what you get.

3. Mar 14, 2004

### mathzeroh

thank you for that awesome reply! ok, i've taken all of the advice. here's what i've got:

27: x=10
$$x= rcos\theta$$
$$10=rcos\theta$$
then, i divided both sides by $$cos\theta$$ and got:
$$\frac{10}{cos\theta}$$
...and since 1/cos theta is the same as sec theta, the answer would be:
r=10sec theta

28: i don't want to make it too long, but my answers that i got were:
r=7 and r=-7

and 29: r=5/2 sin theta

just like in the back of the book! thanks a lot for your help man!!

Last edited: Mar 14, 2004
4. Mar 14, 2004

Uh oh, you were a little careless on #28.

5. Mar 14, 2004

### mathzeroh

what? how?[b(] [b(]

ok here's my work:

x^2+y^2=7
(rcos&theta;)^2+(rsin&theta;)^2=7
r^2(cos&theta;^2+sin&theta;^2)=7
r^2=7
I SEE IT!!
r=THE SQUARE ROOT OF PLUS OR MINUS SEVEN!!!

thanks a lot man!! you're so smart!

6. Mar 14, 2004

Sure, sure.

7. Mar 14, 2004

### mathzeroh

cool thanks a lot

ok. here they are. :

Write each polar equation in rectangular form.
30: r=12
x=rcos&theta;
x=12cos&theta; <---that's my answer for the "x-value"

here's the "y-value":
y=12sin&theta;

is that right?

can you just give me some more advice on these problems, like the ones you did up there? thanks a lot btw for that.

31.&theta;=-45 degrees (sorry i don't know how to make the degree sign)

this is what i did:
x=rcos-45 degrees
x=r(-0.53) <--my answer for the "X-Value"

"y-value":
y=rsin-45 degrees
y=r(-0.85)

y=-x

how??

32. rsin&theta;=4
33. r=-2sec&theta;

thanks agian!!

Last edited: Mar 14, 2004
8. Mar 14, 2004

All right. 30 is technically correct, but it's not what they want. They want you to get rid of theta, too. So try doing this first, square r = 12 and see if you can't figure out what to do. If you're still stuck, think about #28 and how you got that answer, then work backwards.

For 31, what equation did you have relating theta to y and x? You used it somewhere in your first post, so why don't you go find it and see what you can do with it.

32: $r\sin{\theta}$? That looks familiar...

33: Didn't your answer for #27 have a $10\sec{\theta}$? Isn't $-2\sec{\theta}$ pretty similar to $10\sec{\theta}$?

9. Mar 14, 2004

### mathzeroh

it's alright man, i understand. it was pretty late anyway and i had some tough problems there. i noe i was pretty tired, and

you, well it might have been later even than it was here, so thanks a lot for sticking with me that much and that late

man, i appreciate it.

alright. i've taken that wonderful advice again (thank you for taking your time and giving it to me). here's what i've gotten:

30. well as you said, i worked it backwords. at first i was lost, but then i did wut you said and i looked back at

number 28 to see how i did that one, after all, they are pretty dang similar and i'm glad you pointed that out.

working it backwards, i got:

$$r=12$$
$$\sqrt{r^2}=\sqrt{12^2}$$
$$r^2(\cos\theta^2+\sin\theta^2)=144$$
$$(r\cos\theta)^2+(r\sin\theta)^2=144$$
$$x^2+y^2=144$$

and that's the answer, $$x^2+y^2=144$$ <--- because it's in the rectangular form...right?

31. (thanks a lot for reminding me of that formula, i was LOST man! [b(] )
$$\theta=-45\circ$$

the formula:
$$\theta=\tan^{-1}\frac{y}{x}$$

since i have already been given theta, i just plugged it in:
$$-45\circ=\tan^{-1}\frac{y}{x}$$

take the tangent of both sides...
$$-1=\frac{y}{x}$$

multiply both sides by x, to get the answer:
$$-x=y$$

32. well this was pretty obvious too, but i just don't think sometimes
$$r\sin\theta=4$$
rsin&theta; is the same thing as "y"!!
$$r=4$$ DUH!

33.
$$r=-2\sec\theta$$

sec&theta; is the same thing as 1/cos&theta; so i multiplied this to -2 then i multiplied both sides by "cos&theta;"
$$(r=\frac{-2}{\cos\theta})*(\cos\theta)$$
$$r\cos\theta=-2$$
$$x=-2$$

thank you very very much, i really appreciate your help!! thanks a lot cookiemonster!!

Last edited: Mar 14, 2004
10. Mar 29, 2004

### mathzeroh

hello! it's me again! ok i've been working on this problem and its about time that i give it up.

here it is:

Express each equation in polar form. Round "phi" to the nearest degree. ("phi" looks like θ (theta) but

except the line comes down from the top and it is more circular)

y=-1/3x+2

wut i did was bring over the y to that side and i got:
-1/3x-y+2=0

then i went like this:
A=-1/3
B=-1
C=2

i did that to put it into the normal form:
-sqrt((-1/3)^2+(-1)^2) <--negative sqrt because C is positive

i got sqrt(10/9) but i changed that to sqrt(10)/3 because you're not supposed to have a fraction in a sqrt...i think...

then i plugged this into the original formula ---> (-1/3x-y+2=0) to get:
((-1/3)/sqrt(10)/3)x+sqrt((10)/3)y+2sqrt(10)/3=0

i then worked that out to polish it up a bit. this is what i came to:
(3sqrt(10)/10)x+(sqrt10)/3)y+2sqrt(10)/3=0

then i went to solve "phi" which can be solved by this formula:
θ (pretend that is phi, and not theta)=arctan(B/A)
θ=arctan ((sqrt10)/3)/(3/sqrt(10)/10)
that (θ=arctan((sqrt10)/3)/(3/sqrt(10)/10)) would become this:
θ=arctan(10/9) after you clean it up a bit.
=48 degrees

so when i put this into the equation for the polar form, p=rcos(θ-phi)....
i got this:
2sqrt(10)/3=rcos(θ-48) <---theta here is the actual theta

but when after all this work, i cheked in the back, they had this answer:
(3sqrt(10)/5)=rcos(θ-72)

HOW???

Thank you!!

P.S., there might be more from where that came from

Last edited: Mar 29, 2004
11. Mar 29, 2004

### mathzeroh

comeon anybody???

12. Mar 29, 2004

### mathzeroh

well ppl i just came on to say that i have solved the problem

thx for reading this thread though!! i think it was ur guys' plan to make me work on it for hours until i get the answer! thanks very much for your help of not helping me. :tongue: