Fractions and Trig Functions
Date: 11/26/2001 at 14:43:15 From: Mrs. Jabbour Subject: Trig identities In verifying tan2x - cotx = 0, I ran into a problem. If students use (sin2x)/(cos2x) - (cosx)/(sinx) = 0, students get 6 answers, none extraneous. However, if students approach this problem using (2tanx)/(1-tan^2x) = 1/(tanx), they get only 4 of the 6 solutions. What is the reason for using one method over the other in order to assure all possible answers? Is there something I need to tell students about using tan and missing solutions?
Date: 11/26/2001 at 15:44:36 From: Doctor Rob Subject: Re: Trig identities Thanks for writing to Ask Dr. Math, Mrs. Jabbour. This is an interesting phenomenon. The problem is that the identity cot(x) = 1/tan(x) is not valid if x = Pi/2 or 3*Pi/2 (in the range 0 <= x < 2*Pi), because then tan(x) is undefined (or infinite, if you prefer). If one uses this identity, one is implicitly assuming that x does not have these values, and one had better check these two values separately. In the case of the problem at hand, both of these turn out to be solutions, so if one fails to check them, one loses those two solutions. A similar problem occurs with the identity tan(2*x) = 2*tan(x)/(1-tan^2(x)), for the same two values of x. No similar problem occurs with the other approach, because sin(x) and cos(x) are defined for all values of x, so no solutions are lost. Feel free to write again if I can help further. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
Date: 11/26/2001 at 16:19:28 From: Doctor Greenie Subject: Re: Trig identities Hello, Mrs. Jabbour - If a student changes everything into sines and cosines, then he will get something equivalent to cos(x)*[3sin^2(x)-cos^2(x)] --------------------------- = 0 sin(x)*[cos^2(x)-sin^2(x)] which has solutions when the numerator is 0 -- i.e., when either cos(x) = 0 (2 solutions) or 3sin^2(x)-cos^2(x) = 0 (4 solutions) If a student changes tan(2x) and cot(x) into equivalent forms using tan(x), then he will get something equivalent to 3tan^2(x)-1 ------------------- = 0 tan(x)*[1-tan^2(x)] From our usual experience of solving algebraic equations, we think that the only solutions to this equation are when 3tan^2(x)-1=0 (the same 4 solutions obtained above from the equation 3sin^2(x)-cos^2(x)=0) However, there are also solutions when |tan(x)| = infinity (the same 2 solutions obtained above from the equation cos(x)=0) So the "missing" solutions from the students who solve the problem by putting everything in terms of tan(x) are missing only because those students don't realize that a fraction involving trig functions can be 0 if EITHER the numerator is 0 OR if the absolute value of the denominator is infinite. I hope this helps. Write back if you have any further questions about this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum