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

- anonymous

- schrodinger

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- anonymous

Well, let's just draw! :D
|dw:1360754655516:dw|

- anonymous

Cosecant is just hypotenuse over the opposite side, right?

- anonymous

|dw:1360754762915:dw|

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## More answers

- anonymous

no its adj/hyp

- anonymous

Don't mistake cosine for cosecant, they're very different :D

- anonymous

Looked through your notes? Agree with me yet? :D

- anonymous

im so confused. it says in my book COS(theta)=Adj/hyp

- 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

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

And lol @hartnn
Cotangent is positive in the 3rd quadrant...

- anonymous

im stll confused sir or ma'am

- 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

and we need to consider this ...as now the triangle will look like : |dw:1360755574255:dw|

- 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

you gotthat using identities, right ?

- anonymous

i believe so

- 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

ok

- 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

*so, \(\cot \theta = -\sqrt {143}\)

- anonymous

so you use the same equation for cos, sec, tan

- 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

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

Really sorry about disappearing, my internet isn't friendly to me today :(

- anonymous

its ok. i am just glad you helped me. Thank you very much

- anonymous

could you help me with a nother problem im having trouble with?

- anonymous

No problem :)

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