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lilsis76

  • 2 years ago

graph each point in a polar coordinate system, then convert the given polar coordinate to rectangular coordinates. (-1, pi/8) this is all my homework is and i have other ones where i have to reverse the coordinates from recangle to polar

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  1. freewilly922
    • 2 years ago
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    Polar coords give you $$(r,\theta)$$ so that on the axis you draw something like |dw:1351738791260:dw| you can then convert that point to an (x,y) rectangular coord. by using trig and making a triangle. $$y=\sin(\theta) \\ x = \cos(\theta) $$

  2. freewilly922
    • 2 years ago
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    |dw:1351738924218:dw|

  3. lilsis76
    • 2 years ago
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    then where does the pi/8 land? i cant find that

  4. freewilly922
    • 2 years ago
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    radian notation can be converted as follows: 0pi = 0 degrees pi = 180 degrees 2pi = 360 degrees so pi/8 = 180/8 = 22.5 degrees. first quadrant

  5. lilsis76
    • 2 years ago
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    what the heck?! haha

  6. lilsis76
    • 2 years ago
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    i didnt think about the 180 over 8

  7. lilsis76
    • 2 years ago
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    dang it. now im lost :/ i dont know what to do next

  8. freewilly922
    • 2 years ago
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    so to get the (x,y) you use (cos(angle), sin(angle)) so for this one \[\bigg(\cos\left(\frac{\pi}{8}\right),\sin\left(\frac{\pi}{8}\right)\bigg)\]

  9. lilsis76
    • 2 years ago
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    so then ....

  10. lilsis76
    • 2 years ago
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    well i guess since x is -1, x will be -1, does that make the y or theta thingy zero ? (0)?

  11. freewilly922
    • 2 years ago
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    you need to use your calculator for this. why do you think x is -1?

  12. lilsis76
    • 2 years ago
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    dang it, okay its (r, theta) right? okay umm wait. i feel so stupid i dont understand anything im doing

  13. lilsis76
    • 2 years ago
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    well i did the calculaotor thingy u said. cos(pi/8) = the .999 and sin(pi/8)= .007

  14. freewilly922
    • 2 years ago
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    your calculator may be in degree mode instead of radians. Try either switching you mode on the calculator or trying cos(180/8) and sin(180/8)

  15. lilsis76
    • 2 years ago
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    dang.... :/ the back of the book says the answer is\[-\sqrt{2+\sqrt{2}} /2 , -\sqrt{2-\sqrt{2}} /2\]

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