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Zenmo

  • one year ago

Question on Trig. Identities on this problem that involves converting rectangular equation to polar form for xy=16.

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  1. Zenmo
    • one year ago
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    \[xy=16 (1st step), (rcos \theta)(rsin \theta)=16 (2nd step), (\cos \theta)(\sin \theta)r^2=16 (3rd step), \frac{ 1 }{ 2 }(\sin2 \theta)r^2=16 (4th step).\] How do I get from the 3rd step to the 4th step?

  2. anonymous
    • one year ago
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    x = r cos Θ y = r sin Θ xy = 16 (r sin Θ)(r cos Θ) = 16 r² sin Θ cos Θ = 16

  3. zepdrix
    • one year ago
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    From the third step to the 4th? You apply your Sine Double Angle Identity :)

  4. zepdrix
    • one year ago
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    \[\Large\rm \color{orangered}{2\sin \theta \cos \theta=\sin(2\theta)}\]We want to make that show up in our problem. When we're only given this:\[\Large\rm \sin \theta \cos \theta\]We'll make a 2 show up by double two things at once, multiplying by 2, and dividing by 2.\[\Large\rm \frac{1}{2}\cdot2\cdot\sin \theta \cos \theta\]And we can apply our identity from there :)\[\Large\rm \frac{1}{2}\cdot\color{orangered}{2\sin \theta \cos \theta}\]

  5. zepdrix
    • one year ago
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    by doing* two things at once, blah typo

  6. Zenmo
    • one year ago
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    \[(\cos \theta)(\sin \theta)r^2=16\] Could u do the next step for that?

  7. Zenmo
    • one year ago
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    Just a tiny bit confused on the double angle identity part

  8. zepdrix
    • one year ago
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    \[(\cos \theta)(\sin \theta)r^2=16\]\[\sin \theta \cos \theta r^2=16\]\[\frac{1}{2}\cdot2\cdot \sin \theta \cos \theta r^2=16\]\[\frac{1}{2}\cdot\color{orangered}{2\cdot \sin \theta \cos \theta} r^2=16\]\[\frac{1}{2}\cdot\color{orangered}{\sin (2 \theta)} r^2=16\]

  9. Zenmo
    • one year ago
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    Why do we multiply by 2 and divide by 2 for the 2? That is the only part that I don't get, other than that, I understand on how to do rest of the problem.

  10. zepdrix
    • one year ago
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    Well you're probably used to this simple trick in math, do something to one side, do it to the other as well. We're using a different trick, do something to one side, undo it at the same time. Example: \(\Large\rm x= x-2+2\) Maybe I really need an x-2 to show up so I can use it in some way. Like in this example:\[\Large\rm \frac{x}{x-2}\]Yes, we could go through the process of long division, but instead we could add and subtract 2 in the numerator,\[\Large\rm =\frac{x-2+2}{x-2}=\frac{x-2}{x-2}+\frac{2}{x-2}=1+\frac{2}{x-2}\]And that's how we would divide the example I gave.

  11. zepdrix
    • one year ago
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    We're multiplying by 2, we're also dividing by 2 because we have to keep things balanced. Then we `use` the 2 which is multiplying to fix our identity.

  12. Zenmo
    • one year ago
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    Ok, thanks, I got it now after looking at your examples

  13. zepdrix
    • one year ago
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    You can think of it like this if it helps maybe,\[\Large\rm \sin \theta \cos \theta=\frac{2}{2}\sin \theta \cos \theta=\frac{1}{2}\cdot 2\sin \theta \cos \theta\]Ok cool :)

  14. Zenmo
    • one year ago
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    yea, that definitely helps by thinking it as "keeping the balance." :D

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