anonymous
  • anonymous
Find the values of the six trigonometric functions of θ with the given constraint.Function value csc θ = 12, Constraint cot θ < 0
Trigonometry
schrodinger
  • schrodinger
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anonymous
  • anonymous
Well, let's just draw! :D |dw:1360754655516:dw|
anonymous
  • anonymous
Cosecant is just hypotenuse over the opposite side, right?
anonymous
  • anonymous
|dw:1360754762915:dw|

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anonymous
  • anonymous
no its adj/hyp
anonymous
  • anonymous
Don't mistake cosine for cosecant, they're very different :D
anonymous
  • anonymous
Looked through your notes? Agree with me yet? :D
anonymous
  • anonymous
im so confused. it says in my book COS(theta)=Adj/hyp
hartnn
  • hartnn
@PeterPan did u just assume theta here : |dw:1360754965665:dw| because since cot theta is negative, theta is in 3rd quadrant. and so, measure of angle theta is between 180 and 270 degrees.
anonymous
  • anonymous
I drew this triangle with no coordinates in mind... should I have considered them? I was going to reason out that since cotangent is going to be negative, so should tangent, as it's just the reciprocal of tangent. Since cosecant is positive, so should sine, and since tangent, which is just sine over cosine, is negative, so should cosine and secant. :>
anonymous
  • anonymous
And lol @hartnn Cotangent is positive in the 3rd quadrant...
anonymous
  • anonymous
im stll confused sir or ma'am
hartnn
  • hartnn
sorry to confuse you savac , we have , cot as negative, means tan is negative. thats in 2nd or 3rd quadrant, now cosec is positive, means sin is positive, so, angle theta is in 2nd quadrant only. did u get this, first ?
hartnn
  • hartnn
and we need to consider this ...as now the triangle will look like : |dw:1360755574255:dw|
anonymous
  • anonymous
no, i got for my answers sin=1/12, cos=\[\sqrt{143}\] /12 tan=1/\[\sqrt{143}\] csc= 12 sec=12/\[\sqrt{143}\] cot=\[\sqrt{143}\]/1
hartnn
  • hartnn
you gotthat using identities, right ?
anonymous
  • anonymous
i believe so
hartnn
  • hartnn
okk, just make the cos, sec, tan and cot answers as negative. because theta lies in 2nd quadrant as i explained, where all these are negative.
anonymous
  • anonymous
ok
hartnn
  • hartnn
like \(\csc ^2 \theta -1 = \cot^2 \theta \\ \cot^2 \theta = 144-1=143 \\ \cot \theta = \pm \sqrt{143} \) now we select NEGATIVE root , because cot <0 do, same thing for cos, sec, tan
hartnn
  • hartnn
*so, \(\cot \theta = -\sqrt {143}\)
anonymous
  • anonymous
so you use the same equation for cos, sec, tan
hartnn
  • hartnn
same ? no for tan use tan = 1/ cot (since, you have cot) for sin use, sin = 1/csc (since, you have csc) for cos, use cos = tan / sin (since, now u have both, tan and sin) for sec, use sec = 1/ cos (since, now u have cos)
anonymous
  • anonymous
ok so my answers would be sin=1/12 cos=-\[\sqrt{143}\]/12 tan= 1/-\[\sqrt{143}\] csc= 12 sec=12/-\[\sqrt{143}\] cot=-\[\sqrt{143}\]
anonymous
  • anonymous
Really sorry about disappearing, my internet isn't friendly to me today :(
anonymous
  • anonymous
its ok. i am just glad you helped me. Thank you very much
anonymous
  • anonymous
could you help me with a nother problem im having trouble with?
anonymous
  • anonymous
No problem :)

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