anonymous
  • anonymous
Let θ be an angle in quadrant II such that sin
Mathematics
jamiebookeater
  • jamiebookeater
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cwrw238
  • cwrw238
|dw:1372535299105:dw|
cwrw238
  • cwrw238
can you continue from here?
anonymous
  • anonymous
Thanks. I'm not sure what to do from here.

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cwrw238
  • cwrw238
tan x = opposite / adjacent = -1 / sqrt15 sec x = hypotenuse / adjacent = ?
jdoe0001
  • jdoe0001
http://faculty.wlc.edu/buelow/PRC/T-2_1.jpg
jdoe0001
  • jdoe0001
|dw:1372535888225:dw|
anonymous
  • anonymous
Alright now I see. Thanks to both of you.
cwrw238
  • cwrw238
yw
anonymous
  • anonymous
Would the hypotenuse become a negative when calculating sec? Or would it just be: sec x = 4/sqrt15?
anonymous
  • anonymous
Or better yet, why did the 1 become negative when calculating the tangent?
Jhannybean
  • Jhannybean
Because your function is in the second quadrant where sine is positive and cosine is negative.
cwrw238
  • cwrw238
the whole ratio became negative tangent : opposite = 1 , adjacent = - sqrt15 tan = 1 / -sqrt15 = -1/sqrt15
Jhannybean
  • Jhannybean
\[\large \left(-\frac{1}{\sqrt{15}}\right)\]
anonymous
  • anonymous
So, sec = -4/sqrt15?
cwrw238
  • cwrw238
yes
anonymous
  • anonymous
ahh okay I see now thanks.
cwrw238
  • cwrw238
x values to the left of y-axis are negative y values below x axis are also negative

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