The terminal side of #theta# lies on the line #y=1/3x# in quadrant III, how do you find the values of the six trigonometric functions by finding a point on the line?
1 Answer
We're in the third quadrant, where both the x-axis and the y-axis is negative.
I would recommend looking for the first coordinate on the line after the origin where both x and y values are integers. This would be
We now have enough information to determine
#tantheta = "opposite"/"adjacent" = -1/(-3) = 1/3#
#cottheta = 1/tantheta = 1/("opposite"/"adjacent") = "adjacent"/"opposite" = (-3)/(-1) = 3#
We must find the hypotenuse of the triangle to find the other ratios.
#(-3)^2 + (-1)^2 = h^2#
#9 + 1= h^2#
#h = sqrt(10)#
We can now apply the definitions of the other ratios to solve.
#sintheta = "opposite"/"hypotenuse" = -1/sqrt(10) = (-sqrt(10))/10#
#costheta = "adjacent"/"hypotenuse"= -3/sqrt(10) = (-3sqrt(10))/10#
#sectheta = "hypotenuse"/"adjacent" = sqrt(10)/-3 = -sqrt(10)/3#
#csctheta = "hypotenuse"/"opposite" = sqrt(10)/-1 = -sqrt(10)#
Hopefully this helps!
