speedlearner said:
The textbook says as follows: An alternative method of describing the slope of a line with respect to the horizontal or x-axis utilizes the angle which the line makes with the axis. This angle of inclination is the counterclockwise angle whose initial side is the x-axis taken in the positive direction and whose terminal side lies on the line itself taken in the upward direction. Then is shows an illustration.
A few sentences later it says, "The slope of a line and the angle of inclination both give the direction of the line. The slope of a line is m = y2 -y1 / x2 - x1.
When you write this as text on a single line, you
need parentheses, like so:
m = (y2 -y1) / (x2 - x1)
What you wrote would be interpreted as
m = y2 - ##\frac{y1}{x2}## - x1
speedlearner said:
I've used the substitution method of mathematics for many years and I've done a comprehensive study of both Euclid and Descartes. I even studied Ptolomey's table of Chords. But none of these texts mention a reference angle. To this day I have no idea who invented the reference angle or why they invented such a mathematical term. I've asked many people and they are just as dumbfounded as I am.
I'm not sure how useful it is to study "just" geometry, as presented in Euclid and Ptolemy. The combining of the geometry of the Greeks and the algebra of the Arabs and Indians) in analytic geometry was a large step forward in the understanding of mathematics of the time.
With that in mind, it's helpful to think in terms of the unit circle for the trig functions. For a given angle in its standard position, one ray is along the positive x-axis from (0, 0) to (1, 0), and the other ray extends to a point (x, y) on the unit circle. If the angle is acute (lies in Quadrant I), the reference angle is the same as the angle. For an angle such as 135°, for which the terminal ray is in Quadrant II, the terminal ray hits the unit circle at (-√2/2, √2/2). For this angle, the reference angle is the one made by the terminal ray and the negative x-axis, and has a measure of 45°.
All of the trig functions of 135° are the same in absolute value as for 45°, but fairly obviously, some of them are negative in value. For example, sin(135°) and sin(45°) are equal in value, but the cosines (and tangents) of the two angles are opposite in sign.
speedlearner said:
Now in a regular mathematical equation, when you use the substitution method, you have to make sure both sides of the equation are equal.
It's not really a substitution "method." When you substitute one expression for another, you are merely replacing something by something else that has the same value.
speedlearner said:
So when someone presents an example equation like this: 68 = 180 - 112, you could substitute the number 68 for 180 - 112. But in this equation I couldn't just remove the number 180. Otherwise the equation would look like this: 68 = -112. And as we all know 68 does not equal -112. Now can you see my confusion?
Yes. Kline is saying two things when A = 180° - B:
1. tan(A) = tan(180° - B). This part is clear to you, I believe. A and 180° - B are equal expressions, so they both have the same tangent value.
2. tan(A) = -tan(B). This is the part you're having trouble with. As I said in my first post, where this comes from is clear if you have the reference angles to draw on (which Euclid and Ptolemy didn't have).
Let's fall back to my example, with A = 45° and B its supplement (= 135°). Notice that these angles add to 180°, a so-called straight angle.
Angle A
Reference point on unit circle: ( √2/2, √2/2)
Angle B
Reference point on unit circle: ( -√2/2, √2/2)
On the unit circle, the x-coordinate gives the cosine of the angle, and the y-coordinate gives the sine of the angle. The ratio of the y-coordinate over the x-coordinate gives the tangent of the angle, so for this example, tan(A) = 1, and tan(B) = -1 = - tan(A).
It's easy enough to show using ordinary geometry, that this equation holds for arbitrary angles A and B that are supplementary.