To find the values of the trig ratios given a point on the terminal side, we begin by drawing a diagram of the angle in standard position in the coordinate plane. Thus, we place the vertex on the origin, the initial ray as the + x-axis, and draw a terminal ray that for the given Θ will be in Quadrant IV, closer to the x- than the y-axis. Place (6 , -2) on the terminal ray, and draw a perpendicular back to the x-axis.
By doing that, we've created a reference (right) triangle that we can use to relate our answers back to the more familiar right triangle (SOHCAHTOA) trig. Notice that we now have side lengths for the legs of the right triangle, and we can use pythagorean theorem to solve for the hypotenuse (radius). The hypotenuse is √40 = 2√10. Now using either our coordinate plane definitions, or sohcahtoa definitions (using the acute angle made by the terminal ray and the x-axis as the reference angle), we get the following:
sinΘ = -2/2√10 = - √10/10 cscΘ = - √10
cosΘ = 6/2√10 = 3√10/10 secΘ = √10/3
tanΘ = - 1/3 cotΘ = - 3
We can proceed similarly for parts b) and c). In b) we draw a Quadrant II angle whose terminal ray goes thru (- 12 , 5). In c), we are in Quadrant III (the only quadrant in which sine is - but tangent is +), and we have a terminal ray that goes thru the point with y-coordinate = -4 that is 5 units from the origin. By pythag thm again, we get an x-coordinate = - 3.
