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dontothabonbon

Let θ be an angle in quadrant II such that sin

  • 9 months ago
  • 9 months ago

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  1. cwrw238
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    |dw:1372535299105:dw|

    • 9 months ago
  2. cwrw238
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    can you continue from here?

    • 9 months ago
  3. dontothabonbon
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    Thanks. I'm not sure what to do from here.

    • 9 months ago
  4. cwrw238
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    tan x = opposite / adjacent = -1 / sqrt15 sec x = hypotenuse / adjacent = ?

    • 9 months ago
  5. jdoe0001
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    http://faculty.wlc.edu/buelow/PRC/T-2_1.jpg

    • 9 months ago
  6. jdoe0001
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    |dw:1372535888225:dw|

    • 9 months ago
  7. dontothabonbon
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    Alright now I see. Thanks to both of you.

    • 9 months ago
  8. cwrw238
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    yw

    • 9 months ago
  9. dontothabonbon
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    Would the hypotenuse become a negative when calculating sec? Or would it just be: sec x = 4/sqrt15?

    • 9 months ago
  10. dontothabonbon
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    Or better yet, why did the 1 become negative when calculating the tangent?

    • 9 months ago
  11. Jhannybean
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    Because your function is in the second quadrant where sine is positive and cosine is negative.

    • 9 months ago
  12. cwrw238
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    the whole ratio became negative tangent : opposite = 1 , adjacent = - sqrt15 tan = 1 / -sqrt15 = -1/sqrt15

    • 9 months ago
  13. Jhannybean
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    \[\large \left(-\frac{1}{\sqrt{15}}\right)\]

    • 9 months ago
  14. dontothabonbon
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    So, sec = -4/sqrt15?

    • 9 months ago
  15. cwrw238
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    yes

    • 9 months ago
  16. dontothabonbon
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    ahh okay I see now thanks.

    • 9 months ago
  17. cwrw238
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    x values to the left of y-axis are negative y values below x axis are also negative

    • 9 months ago
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