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# FP2: Polar coordinates watch

1. find the polar coordinates for this curve:

just wondering if my working is valid:

im not sure if this is a legit step, square both sides?

i have a huge grudge on that step i took, but just couldnt really think of any alternatives. please help.
2. (Original post by KeineHeldenMehr)
im not sure if this is a legit step, square both sides?
Can you square both sides of x - y = 3? I think you'll find you get a different equation.

(Original post by KeineHeldenMehr)
cos t - sin t = (cos pi/4 cos t - sin pi/4 sin t) * (some constant)...
3. (Original post by generalebriety)
cos t - sin t = (cos pi/4 cos t - sin pi/4 sin t) * (some constant)...
hmm..i dont understand what just happened. all i can figure out is that you've used the double angle formulae for cos2t.
4. (Original post by KeineHeldenMehr)
hmm..i dont understand what just happened. all i can figure out is that you've used the double angle formulae for cos2t.
Yeah, I'm not finding any particularly easy way to explain it. Well, I know that cos pi/4 = sin pi/4, and I knew I wanted to crowbar a double angle formula into there somewhere, so I just multiplied through by that common value and then divided by it (which is the "some constant"), and you'll then be able to combine what you get using the double angle formula.

There's probably a more A-level-friendly way to do it, but I'm not seeing it. Doesn't mean it doesn't exist...
5. You know how in C3 you learn that Asinx + Bcosx = rcos(x + alpha) by expanding the RHS? Well over here, express sin(theta) - cos(theta) in the form kcos(theta + alpha) and you'll know what k and alpha are.

So then you'll have r(sin(theta) - cos(theta) = r(kcos(theta + alpha)) = rkcos(theta + alpha) = 3

Judging by their answer, k = root2 and alpha = pi/4 (and thinking about it, that sounds right as well).
6. (Original post by Swayum)
You know how in C3 you learn that Asinx + Bcosx = rcos(x + alpha) by expanding the RHS? Well over here, express sin(theta) - cos(theta) in the form kcos(theta + alpha) and you'll know what k and alpha are.

So then you'll have r(sin(theta) - cos(theta) = r(kcos(theta + alpha)) = rkcos(theta + alpha) = 3

Judging by their answer, k = root2 and alpha = pi/4 (and thinking about it, that sounds right as well).
hmm..now that C3 is mentioned, things sort of make sense..

i dont fully understand this jump:

r[cos(theta)-sin(theta)]=3

r cos(theta + alpha)

beyond this point i understand everything else. since i get

1=r cos(alpha)
1=r sin(alpha)

where i can get alpha and r.

your use of k has rather puzzled me too.
7. The r you're using in x = rcos(theta) and y = rsin(theta) is different to the r I was referring to in the C3 identity "Asinx + Bcosx = rcos(x + alpha)". Therefore, I'm changing the C3 identity to "Asinx + Bcosx = kcos(x + alpha)".

Literally all you need to do is express sin(theta) - cos(theta) as one trig term.

Then just sub it in.
8. (Original post by Swayum)
The r you're using in x = rcos(theta) and y = rsin(theta) is different to the r I was referring to in the C3 identity "Asinx + Bcosx = rcos(x + alpha)". Therefore, I'm changing the C3 identity to "Asinx + Bcosx = kcos(x + alpha)".

Literally all you need to do is express sin(theta) - cos(theta) as one trig term.

Then just sub it in.
ah i see. thanks.
9. That's so weird - I literally did this question this afternoon! I got the same answer as you and my teacher said it was right, it was just in a different form to the answer in the book.

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Updated: February 3, 2010
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