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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well, let's just draw! :D dw:1360754655516:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Cosecant is just hypotenuse over the opposite side, right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1360754762915:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Don't mistake cosine for cosecant, they're very different :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Looked through your notes? Agree with me yet? :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0im so confused. it says in my book COS(theta)=Adj/hyp

hartnn
 3 years ago
Best 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.0And lol @hartnn Cotangent is positive in the 3rd quadrant...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0im stll confused sir or ma'am

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0sorry 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
 3 years ago
Best ResponseYou've already chosen the best response.0and we need to consider this ...as now the triangle will look like : dw:1360755574255:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no, 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
 3 years ago
Best ResponseYou've already chosen the best response.0you gotthat using identities, right ?

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0okk, 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.

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

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0*so, \(\cot \theta = \sqrt {143}\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so you use the same equation for cos, sec, tan

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0same ? 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
 3 years ago
Best ResponseYou've already chosen the best response.0ok 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
 3 years ago
Best ResponseYou've already chosen the best response.0Really sorry about disappearing, my internet isn't friendly to me today :(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0its ok. i am just glad you helped me. Thank you very much

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0could you help me with a nother problem im having trouble with?
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