Solving Trigonometric Equations in RadiansDate: 12/14/97 at 00:26:48 From: alex r Subject: Solving trigonometric equations in radians Solve for theta in radians to 2 decimals if 0 < theta < 2 pi tan theta = -0.318. I tried inverse tan -0.318 (in radians mode) and it gave me -0.307; however, it said between 0 and 2pi so I added 2pi and got one of the answers, 5.9. However, I can only get the other 2.8 if I add 1 pi. Why am I adding one pi? Whenever I do something in degrees, if I am given an angle and want to find another angle with the same tan or whatever, I add 360. Here pi = 180 - why am I adding pi (180) and not 2pi (360)? Please help. Thanks. Date: 12/14/97 at 04:28:51 From: Doctor Luis Subject: Re: Solving trigonometric equations in radians This is a good question. To understand why you only add one pi, you need to understand something about a special property of trigonometric functions called periodicity. As you may know, trigonometric functions are periodic. A periodic function f(x) has the following special property, f(x+P) = f(x) This means that the function repeats itself (it gives you the same values over again) when you add a special number. Since there's an infinite number of such numbers, we usually take the smallest positive number such that the equation above is still true, and we call that number the period of the function f(x). let p = period of f(x). You can see that if you add p to x a lot of times you'll still get the same value f(x+np) = f(x) , where n is an integer (negative, positive or zero) This gives you a problem if you want to solve some equation like the following for x (there's an infinite number of values of x that will give you the same value of y!) f(x) = y ----> x + np = inversef(y) or x = inversef(y) + mp , where m = -n Since you know that you have an infinite values of x, and you also know the period of f(x), we usually take the solution x with m = 0, so that x = inversef(y) Notice that, in this case, x will have to be between 0 and p, because if it were greater than p you could always subtract p until you got x in between 0 and p, or if x were less than 0 you could always add p until you got x in between 0 and p. Then, 0 < x < p The solution with this restriction on x is called the principal solution of the periodic function f(x) = y You are already familiar with the trigonometric functions sine and cosine: their period is the same and is equal to 2pi radians or 360 degrees. In fact, the cosine is the same curve as the sine, except that it is shifted pi/2 radians to the right (that's why they have the same period) sin(x+2pi) = sin(x) and cos(x+2pi) = cos(x) Now, the tangent doesn't have a period of 2pi like the sine and cosine. It has a period of 1 pi. This is because of the way the tangent is defined, sin(x) tan(x) = -------- cos(x) So, sin(x+pi) tan(x+pi) = ----------- cos(x+pi) Since the cosine is just a shifted sine curve, cos((pi/2)-x) = sin(x) This is the same as saying that sin((pi/2)-x) = cos(x) Now, you can rewrite sin(x+pi) and cos(x+pi) like this, sin(x+pi) = sin(pi+x) = sin((pi/2)-(-x-(pi/2))) = cos(-x-(pi/2)) = cos(-((pi/2)+x)) = cos((pi/2)+x) (b/c cos(-x) = cos(x) ) = cos((pi/2)-(-x)) = sin(-x) = -sin(x) (b/c sin(-x) = sin(x) ) and, cos(x+pi) = cos(pi+x) = cos((pi/2)-(x-(pi/2))) = sin(x-(pi/2)) = -sin((pi/2)-x) (b/c sin(-x) = sin(x) ) = -cos(x) Then, sin(x+pi) tan(x+pi) = ----------- cos(x+pi) - sin(x) = ----------- - cos(x) sin(x) = -------- cos(x) = tan(x) (that's just the definition!) And so, you see tan(x+pi) = tan(x) that's just a periodic function with period pi !! Now, your problem said: solve for x in tan(x) = -0.318 with 0 < x < 2pi so, x = -0.307 (radians) but remember, our solution for a periodic function has to be between 0 and the period p. In this case, we just proved that the period p of the tan(x) function is pi (radians). So we add pi to x, and get x + pi = 2.83 (radians) This brings our solution to be between 0 and pi (3.14...) and gives us the principal solution of the equation tan(x) = -0.318 Remember that a periodic function has an infinite number of solutions? Since the problem asks for solutions between 0 and 2 pi there are going to be two solutions. One between 0 and pi, and the other one between pi and pi+pi=2pi. So, to get the other solution we just add pi (pi is the period, remember?) to our principal solution, and get (x+pi)+pi = 2.83 + pi = 5.97 (radians) If the problem had asked for a solution with 0 < x < 3pi, there would have been three solutions, and we would have had to add pi (the period of tan(x)) again, (x+2pi)+pi = 5.97 + pi = 9.11 (radians) If the problem had asked for a solution between -pi and 0, we would have kept the solution your calculator gave you in the first place. x = invtan(-0.318) = -0.307 So, you see, knowing the principal solution and the period of a periodic function gives you all of the solutions. All you have to do is find the solutions that you want (for example, all the solutions between -10 and 345). What if the problem had said, instead: solve for x, tan(2x) = -0.318 with 0 < x < 2pi what would the solutions be? (hint: there are four solutions) Something to think about: If a periodic function f(x) has period p, what is the period of the periodic function f(rx+s), where r is any positive real number and s any real number? I hope this helped :) -Doctor Luis, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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