Question 1203393
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I'll use theta in place of t


cos(theta) = -2/3
cos(theta) = adjacent/hypotenuse


adjacent = -2
hypotenuse = 3


Use the pythagorean theorem {{{a^2+b^2=c^2}}} to determine the opposite side is {{{sqrt(5)}}}. 
I'll make this value negative to indicate we're below the x axis.


adjacent = -2
hypotenuse = 3
opposite = {{{-sqrt(5)}}}


Here's what the diagram looks like
{{{drawing(400,400,-5,5,-5,5,
line(-5-3,0,5+3,0),
line(0,-5-3,0,5+3),
red(line(0,0,-3,-4)),
red(line(-3,-4,-3,0)),
red(line(-3,0,0,0)),
locate(-1,-0.5,theta),

locate(-4,-2,"Opp"),
locate(-2,-0.1,"Adj"),
locate(-1.5,-2.4,"Hyp"),

locate(-4,-1.5,-sqrt(5)),
locate(-2,0.6,"-2"),
locate(-1,-1.9,"3"),

locate(5-0.4,5-0.4,"Q1"),
locate(-4.6,5-0.4,"Q2"),
locate(-4.6,-4.6,"Q3"),
locate(5-0.4,-4.6,"Q4"),
locate(0.3,-4,matrix(1,2,"Diagram","not")),
locate(0.3,-4.4,matrix(1,2,"to","scale"))
)}}}


Then,
{{{sin(theta) = opposite/hypotenuse = -sqrt(5)/3}}}


{{{tan(theta) = opposite/adjacent = (-sqrt(5))/(-2) = sqrt(5)/2}}}


{{{csc(theta) = hypotenuse/opposite = 3/(-sqrt(5)) = -3*sqrt(5)/5}}} (this is the reciprocal of sine)


{{{sec(theta) = hypotenuse/adjacent = (3)/(-2) = -3/2}}} (this is the reciprocal of cosine)


{{{cot(theta) = adjacent/opposite = (-2)/(-sqrt(5)) = 2/(sqrt(5)) = 2sqrt(5)/5}}} (this is the reciprocal of tangent)


A similar question is found here:
<a href="https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq.question.1203332.html">https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq.question.1203332.html</a>
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