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dontothabonbon

  • one year ago

Let θ be an angle in quadrant II such that sin

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  1. cwrw238
    • one year ago
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    |dw:1372535299105:dw|

  2. cwrw238
    • one year ago
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    can you continue from here?

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

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

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

  6. jdoe0001
    • one year ago
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    |dw:1372535888225:dw|

  7. dontothabonbon
    • one year ago
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    Alright now I see. Thanks to both of you.

  8. cwrw238
    • one year ago
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    yw

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

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

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

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

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

  14. dontothabonbon
    • one year ago
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    So, sec = -4/sqrt15?

  15. cwrw238
    • one year ago
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    yes

  16. dontothabonbon
    • one year ago
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    ahh okay I see now thanks.

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

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