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### 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/
```
Associated Topics:
High School Trigonometry

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