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Find the values of the six trigonometric functions of θ with the given constraint.Function value csc θ = 12, Constraint cot θ < 0
 one year ago
 one year ago
Find the values of the six trigonometric functions of θ with the given constraint.Function value csc θ = 12, Constraint cot θ < 0
 one year ago
 one year ago

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PeterPanBest ResponseYou've already chosen the best response.1
Well, let's just draw! :D dw:1360754655516:dw
 one year ago

PeterPanBest ResponseYou've already chosen the best response.1
Cosecant is just hypotenuse over the opposite side, right?
 one year ago

PeterPanBest ResponseYou've already chosen the best response.1
dw:1360754762915:dw
 one year ago

PeterPanBest ResponseYou've already chosen the best response.1
Don't mistake cosine for cosecant, they're very different :D
 one year ago

PeterPanBest ResponseYou've already chosen the best response.1
Looked through your notes? Agree with me yet? :D
 one year ago

savacBest ResponseYou've already chosen the best response.0
im so confused. it says in my book COS(theta)=Adj/hyp
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
@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.
 one year ago

PeterPanBest ResponseYou've already chosen the best response.1
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. :>
 one year ago

PeterPanBest ResponseYou've already chosen the best response.1
And lol @hartnn Cotangent is positive in the 3rd quadrant...
 one year ago

savacBest ResponseYou've already chosen the best response.0
im stll confused sir or ma'am
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
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 ?
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
and we need to consider this ...as now the triangle will look like : dw:1360755574255:dw
 one year ago

savacBest ResponseYou've already chosen the best response.0
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
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
you gotthat using identities, right ?
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
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.
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
like \(\csc ^2 \theta 1 = \cot^2 \theta \\ \cot^2 \theta = 1441=143 \\ \cot \theta = \pm \sqrt{143} \) now we select NEGATIVE root , because cot <0 do, same thing for cos, sec, tan
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
*so, \(\cot \theta = \sqrt {143}\)
 one year ago

savacBest ResponseYou've already chosen the best response.0
so you use the same equation for cos, sec, tan
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
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)
 one year ago

savacBest ResponseYou've already chosen the best response.0
ok so my answers would be sin=1/12 cos=\[\sqrt{143}\]/12 tan= 1/\[\sqrt{143}\] csc= 12 sec=12/\[\sqrt{143}\] cot=\[\sqrt{143}\]
 one year ago

PeterPanBest ResponseYou've already chosen the best response.1
Really sorry about disappearing, my internet isn't friendly to me today :(
 one year ago

savacBest ResponseYou've already chosen the best response.0
its ok. i am just glad you helped me. Thank you very much
 one year ago

savacBest ResponseYou've already chosen the best response.0
could you help me with a nother problem im having trouble with?
 one year ago
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