
Quadrants and Angles (page 1 of 3)
Sections: Introduction, Worked Examples (and Sign Chart), More Examples
You've worked with trig ratios in a geometrical context: the context of right triangles. Now we'll move those ratios into an algebraic context, and then we'll dispense with the triangles.
Each angle of this triangle measures 60°.


Then let's drop the perpendicular, splitting the triangle into two 306090 triangles: We'll work with that 60° angle, as labelled.







Since the length of the base is the same as the value of "adjacent", we can call this just "x". Similarly, the "opposite" is now just "y".
The Pythagorean Theorem gives us the value of the hypotenuse, which we will now call "r". Why? Because we can also look at this distance along the terminal side of the angle as being a radius line of a circle. Any right triangle with the samelength hypotenuse will have its far end lying on this circle. We already know that cos(60°) = 1/2. Now you can read this value from the drawing in the quadrants. The cosine ratio had been "adjacent over hypotenuse"; now it is "x over r". But the value is still 1/2. 

So we've set up the angle in the first quadrant, thrown away the triangle, and still been able to find the trig ratio. Can we build on this? Yes!
Now grab your calculator, make sure it's in "degree" mode, and plug in "cos(120°)". Did you get "–0.5"? That's the same as –1/2, right? Now look again at the graph: x/r
= (–1)/2 = –1/2 

Does this continue to work in the other quadrants? You bet!
Draw the corresponding line in QIII, marking 240°, being 60° below the negative xaxis. You can see from the picture that cos(240°) ought to be –1/2. What does your calculator say? 


Now move to QIV and draw the line for 300°, being 60° below the positive xaxis. In QIV, x is again positive, so cos(300°) = x/r = 1/2 = 0.5. What does your calculator say? 

Note that in each case, all that really mattered was the sixtydegree triangle in the quadrant where the angle terminated, and the x and yvalues in that quadrant. The fact that you can't draw a right triangle with a base angle of 300° was irrelevant. To find the trig ratios, you needed only to draw the terminal side of the angle in whatever quadrant it happened to land, construct a right triangle by "dropping a perpendicular" to the xaxis, and then work from there. This works for all angles....
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