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anonymous

  • one year ago

medal award!

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  1. anonymous
    • one year ago
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  2. Mertsj
    • one year ago
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    |dw:1444010001441:dw|

  3. Mertsj
    • one year ago
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    |dw:1444010141884:dw|

  4. anonymous
    • one year ago
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    I don't really understand what is what..

  5. Mertsj
    • one year ago
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    It's learning the unit circle I think with emphasis on the angles of the special right triangles.

  6. anonymous
    • one year ago
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    if i'm not mistaken the two distances than are 5pi/4 and pi/4?

  7. Mertsj
    • one year ago
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    \[\sin \frac{\pi}{4}=\cos \frac{\pi}{4}=\frac{\sqrt{2}}{2}\]

  8. Mertsj
    • one year ago
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    So whenever you have a reference angle of pi/4, the sin and cos will be sqrt2/2 making adjustment for the sign depending on the quadrant.

  9. Mertsj
    • one year ago
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    So if the cos, which is like x, is -sqrt2/2, you must be in either quadrant 2 or 3 and the reference angle must be pi/4

  10. Mertsj
    • one year ago
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    And so the distance on the unit circle would be pi-pi/4 and pi + pi/4 or 3pi/4 and 5pi/4

  11. anonymous
    • one year ago
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    but speficially for this problem when it refers to the distances they just want (sqrt(2)/2,sqrt(2)/2)?

  12. Mertsj
    • one year ago
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    No. They want the distance traveled on the unit circle.

  13. Mertsj
    • one year ago
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    Or in other words, they want the angle

  14. anonymous
    • one year ago
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    I put in 5pi/4 and 3pi/4 it said it's incorrect

  15. Mertsj
    • one year ago
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    Well then it must be a formatting problem because if the x value is -sqrt2/2, the angles are 3pi/4 and 5pi/4

  16. Mertsj
    • one year ago
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    did you put in the comma?

  17. anonymous
    • one year ago
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    i put in 3pi/4, 5pi/4

  18. anonymous
    • one year ago
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    it's right! I probably forgot to add in the comma the first time :) thank you

  19. Mertsj
    • one year ago
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    yw

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