Trig without Tears Part 4:
Trig without Tears Part 4:
Summary: The six trig functions were originally defined for acute angles in triangles, but now we define them for any angle (or any number). If you want any of the six function values for an angle that’s not between 0 and 90° (π/2), you just find the function value for the reference angle that is within that interval, and then possibly apply a minus sign.
So far we have defined the six trig functions as ratios of sides of a right triangle. In a right triangle, the other two angles must be less than 90°, as suggested by the picture at left.
Suppose you draw the triangle in a circle this way, with angle A at the origin and the circle’s radius equal to the hypotenuse of the triangle. The hypotenuse ends at the point on the circle with coordinates (x,y), where x and y are the lengths of the two legs of the triangle. Then using the standard definitions of the trig functions, you have
sin A = opposite/hypotenuse = y/r
cos A = adjacent/hypotenuse = x/r
This is the key to extending the trig functions to any angle.
The trig functions had their roots in measuring sides of triangles, and chords of a circle (which is practically the same thing). If we think about an angle in a circle, we can extend the trig functions to work for any angle.
In the diagram, the general angle A is drawn in standard position, just as we did above for an acute angle. Just as before, its vertex is at the origin and its initial side lies along the positive x axis. The point where the terminal side of the angle cuts the circle is labeled (x,y).
(This particular angle happens to be between 90° and 180° (π/2 and π), and we say it lies in Quadrant II. But you could draw a similar diagram for any angle, even a negative angle or one >360°.)
Now let’s define sine and cosine of angle A, in terms of the coordinates (x,y) and the radius r of the circle:
(21) sin A = y/r, cos A = x/r
This is nothing new. As you saw above when A was in Quadrant I, this is exactly the definition you already know from equation 1: sin A = opposite/hypotenuse, cos A = adjacent/hypotenuse. We’re just extending it to work for any angle.
The other function definitions don’t change at all. From equation 3 we still have
tan A = sin A / cos A
which means that
tan A = y/x
and the other three functions are still defined as reciprocals (equation 5).
Once again, there’s nothing new here: we’ve just extended the original definitions to a larger domain.
So why go through this? Well, for openers, not every triangle is an acute triangle. Some have an angle greater than 90°. Even in down-to-earth physical triangles, you’ll have to be concerned with functions of angles greater than 90°.
Beyond that, it turns out that all kinds of physical processes vary in terms of sines and cosines as functions of time: height of the tide; length of the day over the course of a year; vibrations of a spring, or of atoms, or of electrons in atoms; voltage and current in an AC circuit; pressure of sound waves, Nearly every periodic process can be described in terms of sines and cosines.
And that leads to a subtle shift of emphasis. You started out thinking of trig functions of angles, but really the domain of trig functions is all real numbers, just like most other functions. How can this be? Well, when you think of an “angle” of so-and-so many radians, actually that’s just a pure number. For instance, 30°=π/6. We customarily say “radians” just to distinguish from degrees, but really π/6 is a pure number. When you take sin(π/6), you’re actually evaluating the function sin(x) at x = π/6 (about 0.52), even though traditionally you’re taught to think of π/6 as an angle.
We won’t get too far into that in these pages, but here’s an example. If the average water depth is 8 ft in a certain harbor, and the tide varies by ±3 ft, then the height at time t is given by a function that resembles y = 8 + 3 cos(0.52t). (It’s actually more complicated, because high tides don’t come at the same time every day, but that’s the idea.)
Coming back from philosophy to the nitty-gritty of computation, how do we find the value of a function when the angle (or number) is outside the range [0;90°] (which is 0 to π/2)? The key is to define a reference angle.
Here’s the same picture of angle A again, but with its reference angle added. With angle A in standard position, the reference angle is the acute angle between the terminal side of A and the positive or negative x axis. In this case, angle A is in Q II, the reference angle is 180°−A (π−A). Why? Because the two angles together equal 180° (π).
What good does the reference angle do you? Simply this: the six function values for any angle equal the function values for its reference angle, give or take a minus sign.
That’s an incredibly powerful statement, if you think about it. In the drawing, A is about 150° and the reference angle is therefore about 30°. Let’s say they’re exactly 150° and 30°, just for discussion. Then sine, cosine, tangent, cotangent, secant, and cosecant of 150° are equal to those same functions of 30°, give or take a minus sign.
What’s this “give or take” business? That’s what the next section is about.
Remember the extended definitions from equation 21:
sin A = y/r, cos A = x/r
The radius r is always taken as positive, and therefore the signs of sine and cosine are the same as the signs of y and x. But you know which quadrants have positive or negative y and x, so you know for which angles (or numbers) the sine and cosine are positive or negative. And since the other functions are defined in terms of the sine and cosine, you also know where they are positive or negative.
Spend a few minutes thinking about it, and draw some sketches. For instance, is cos 300° positive or negative? Answer: 300° is in Q IV, which is in the right-hand half of the circle. Therefore x is positive, and the cosine must be positive as well. The reference angle is 60° (draw it!), so cos 300° equals cos 60° and not −cos 60°.
You can check your thinking against the chart that follows. Whatever you do, don’t memorize the chart! Its purpose is to show you how to reason out the signs of the function values whenever you need them, not to make you waste storage space in your brain.
|Signs of Function Values|
0 to 90°
0 to π/2
90 to 180°
π/2 to π
180 to 270°
π to 3π/2
270 to 360°
3π/2 to 2π
What about other angles? Well, 420° = 360°+60°, and therefore 420° ends in the same position in the circle as 60°—it’s just going once around the circle and then an additional 60°. So 420° is in Q I, just like 60°.
You can analyze negative angles the same way. Take −45°. That occupies the same place on the circle as +315° (360°−45°). −45° is in Q IV.
As you’ve seen, for any function you get the numeric value by considering the reference angle and the positive or negative sign by looking where the angle is.
Example: What’s cos 240°? Solution: Draw the angle and see that the reference angle is 60°; remember that the reference angle always goes to the x axis, even if the y axis is closer. cos 60° = ½, and therefore cos 240° will be ½, give or take a minus sign. The angle is in Q III, where x is negative, and therefore cos 240° is negative. Answer: cos 240° = −½.
Example: What’s tan(−225°)? Solution: Draw the angle and find the reference angle of 45°. tan 45° = 1. But −225° is in Q II, where x is negative and y is positive; therefore y/x is negative. Answer: tan(−225°) = −1.
The techniques we worked out above can be generalized into a set of identities. For instance, if two angles are supplements then you can write one as A and the other as 180°−A. You know that one will be in Q I and the other in Q II, and you also know that one will be the reference angle of the other. Therefore you know at once that the sines of the two angles will be equal, and the cosines of the two will be numerically equal but have opposite signs.
This diagram may help:
Here you see a unit circle (r = 1) with four identical triangles. Their angles A are at the origin, arranged so that they’re mirror images of each other, and their hypotenuses form radii of the unit circle. Look at the triangle in Quadrant I. Since its hypotenuse is 1, its other two sides are cos A and sin A.
The other three triangles are the same size as the first so their sides must be the same length as the sides of the first triangle. But you can also look at the other three radii as belonging to angles 180°−A in Quadrant II, 180°+A in Quadrant III, and −A or 360°−A in Quadrant IV. All the others have a reference angle equal to A. From the symmetry, you can immediately see things like sin(180°+A) = −sin A and cos(−A) = cos A.
The relations are summarized below. Don’t memorize them! Just draw a diagram whenever you need them—it’s easiest if you use a hypotenuse of 1. Soon you’ll find that you can quickly visualize the triangles in your mind and you won’t even need to draw a diagram. The identities for tangent are easy to derive: just divide sine by cosine as usual.
|sin(180°−A) = sin A
sin(π−A) = sin A
|cos(180°−A) = −cos A
cos(π−A) = −cos A
|tan(180°−A) = −tan A
tan(π−A) = −tan A
|sin(180°+A) = −sin A
sin(π+A) = −sin A
|cos(180°+A) = −cos A
cos(π+A) = −cos A
|tan(180°+A) = tan A
tan(π+A) = tan A
|sin(−A) = −sin A||cos(−A) = cos A||tan(−A) = −tan A|
You should be able to see that 360° brings you all the way around the circle. That means that an angle of 360°+A or 2π+A is the same as angle A. Therefore the function values are unchanged when you add 360° or a multiple of 360° (or 2π or a multiple) to the angle. Also, if you move in the opposite direction for angle A, that’s the same angle as 360°−A or 2π−A, so the function values of −A and 360°−A (or 2π−A) are the same.
For this reason we say that sine and cosine are periodic functions with a period of 360° or 2π. Their values repeat over and over again. Of course secant and cosecant, being reciprocals of cosine and sine, must have the same period.
What about tangent and cotangent? They are periodic too, but their period is 180° or π: they repeat twice as fast as the others. You can see this from equation 22: tan(180°+A) = tan A says that the function values repeat every 180°.
next: 5/Solving Triangles