Question 1202445
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Answers:
{{{sin(theta) = (-2*sqrt(13))/13}}}
{{{cos(theta) = (3*sqrt(13))/13}}}
{{{tan(theta) = -2/3}}}
{{{csc(theta) = (-sqrt(13))/2}}}
{{{sec(theta) = (sqrt(13))/3}}}
{{{cot(theta) = -3/2}}}



Explanation:


The soh cah toa formulas are
{{{sin(theta) = opposite/hypotenuse}}}
{{{cos(theta) = adjacent/hypotenuse}}}
{{{tan(theta) = opposite/adjacent}}}
A shorter format would be
{{{sin(theta) = y/r}}}
{{{cos(theta) = x/r}}}
{{{tan(theta) = y/x}}}
This shorter format is very useful when we're given coordinates of the terminal point.


In this case we have (x,y) = (6,-4) that is the terminal point.
Use the pythagorean theorem to find the radius of the circle, which will give the hypotenuse of the right triangle.
{{{a^2+b^2 = c^2}}}
{{{x^2+y^2 = r^2}}}
{{{r = sqrt(x^2+y^2)}}}
{{{r = sqrt(6^2+(-4)^2)}}}
{{{r = sqrt(52)}}}
{{{r = sqrt(4*13)}}}
{{{r = sqrt(4)*sqrt(13)}}}
{{{r = 2*sqrt(13)}}}


Then,
{{{sin(theta) = y/r = -4/(2*sqrt(13)) = -2/(sqrt(13)) = (-2*sqrt(13))/13}}}
{{{cos(theta) = x/r = 6/(2*sqrt(13)) = 3/(sqrt(13)) = (3*sqrt(13))/13}}}
{{{tan(theta) = y/x = -4/6 = -2/3}}}


The other three trig ratios are the reciprocals of those previously mentioned three ratios.
csc = 1/sin
sec = 1/cos
cot = 1/tan
meaning that
csc = r/y
sec = r/x
cot = x/y
{{{csc(theta) = r/y = (2*sqrt(13))/(-4) = (-sqrt(13))/2}}}
{{{sec(theta) = r/x = (2*sqrt(13))/6 = (sqrt(13))/3}}}
{{{cot(theta) = x/y = 6/(-4) = -3/2}}}
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