Trig Functions, Reference Angles, and the Unit Circle
Date: 07/02/2004 at 11:22:09 From: Karen Subject: Trig Functions & Angles greater than 90 Degrees Hello Dr. Math, I'm confused with the trigonometric functions. Why is a reference angle used to calculate an angle greater than 90 degrees? From the definition of the trigonometric functions, they are used for acute angles. If an angle is greater than 90 degrees, how is a reference angle used? Thanks in advance. Karen
Date: 07/02/2004 at 13:00:51 From: Doctor Peterson Subject: Re: Trig Functions & Angles greater than 90 Degrees Hi, Karen. Trigonometry can be taught in several different ways, depending on the order in which the teacher wants to introduce concepts. It sounds like your class has started with the right triangle version, and is now moving into working with angles greater than 90 degrees, which requires extending those ideas. This can be a little misleading, because it suggests that trig functions are defined only for acute angles, and then you have to use some tricks to find functions of larger angles. What I prefer is to motivate the idea of the trig functions using right triangles, but quickly extend that idea so that our actual definition of the trig functions is based on the "unit circle" concept. Then there is no trouble defining the functions for any angle; but you find that there are rules that allow you to find, say, the sine of a large angle in terms of the sine of a specific acute angle, called its reference angle. Let's take the sine. In a right triangle, this is the "opposite over the hypotenuse": + c / | / |a sin(A) = a/c /A | +-----------+ b If we take this same angle and put it at the origin of a coordinate plane, and then change c to 1 (making a similar triangle), we have ooooooooooo ooo | ooo oo | oo o | o o | + (x,y) o | / |o o | / |yo o | / A | o o-------------+-----------+-o x Now the opposite side is just y, the y-coordinate of the point on the unit circle at the given angle A. The hypotenuse is 1, so the sine of angle A is y/1 or y. Now we extend the idea: for ANY angle A, define the sine of that angle to be the y-coordinate of the corresponding point on the unit circle. How, then, are sines of non-acute angles related to those of acute angles? For angles in the second quadrant, we find that the sine is the same as that of the supplementary angle, which is acute: ooooooooooo ooo | ooo oo | oo o | o (x,y) +-----------+-----------+ o| \ | |o o | \ | |yo o | \ B | o o-+-----------+-----------+-o x Here B is the angle from the positive x-axis to our point (x,y), which is about 150 degrees. But the sine is the same as that of our original angle (30 degrees), because both points are the same height above the x-axis. The rule is sin(180 - A) = sin(A) You'll also notice that the cosine of our angle (x) is negative here, but has the same absolute value as that of the original angle; so cos(180 - A) = -cos(A) Other rules are: sin(-A) = -sin(A) cos(-A) = cos(A) sin(A + 360) = sin(A) cos(A + 360) = cos(A) These rules allow you to express the sine or cosine of any angle as equal to, or the negative of, some acute angle. Here are some pages that may help: Signs of Sines (and Other Trigonometric Ratios) http://mathforum.org/library/drmath/view/60938.html The Unit Circle http://mathforum.org/library/drmath/view/53944.html If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 07/02/2004 at 15:34:00 From: Karen Subject: Thank you (Trig Functions & Angles greater than 90 Degrees) That demystifies a lot. The fact that the sine of 30 and 150 degrees are the same, because those angles have the same y-coordinate on the unit circle, explains it all. Thank you very much, Dr. Math!
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