Here's the question you clicked on:
savac
Find the values of the six trigonometric functions of θ with the given constraint.Function value csc θ = 12, Constraint cot θ < 0
Well, let's just draw! :D |dw:1360754655516:dw|
Cosecant is just hypotenuse over the opposite side, right?
Don't mistake cosine for cosecant, they're very different :D
Looked through your notes? Agree with me yet? :D
im so confused. it says in my book COS(theta)=Adj/hyp
@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.
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. :>
And lol @hartnn Cotangent is positive in the 3rd quadrant...
im stll confused sir or ma'am
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 ?
and we need to consider this ...as now the triangle will look like : |dw:1360755574255:dw|
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
you gotthat using identities, right ?
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.
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
*so, \(\cot \theta = -\sqrt {143}\)
so you use the same equation for cos, sec, tan
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)
ok so my answers would be sin=1/12 cos=-\[\sqrt{143}\]/12 tan= 1/-\[\sqrt{143}\] csc= 12 sec=12/-\[\sqrt{143}\] cot=-\[\sqrt{143}\]
Really sorry about disappearing, my internet isn't friendly to me today :(
its ok. i am just glad you helped me. Thank you very much
could you help me with a nother problem im having trouble with?