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Extraneous Roots Introduced Trigonometrically

Date: 06/10/2012 at 10:50:41
From: Hester
Subject: Losing or introducing roots in a (trigonometric) equation

When proving trigonometric identities or solving trig problems that
involve more than one function, it's common to divide through by a
function, or even multiply through by a function, in order to make the
equation solvable by having just one function of x.

For example, when proving that sec^2(x) = 1 + tan^2(x), one way is to
divide the equation sin^2(x) + cos^2(x) = 1 by cos^2(x), and then
simplify. Therefore, it is obvious that in some situations division is

However, in some equations, when dividing through by a function you lose
one of the roots of the equation. For example, sin^2(t) = cos(t)
simplifies to cos(t)(2sin(t) - 1) = 0. But in this case, if you were to
divide through by cos(t), then you would lose a solution, which would be
cos(t) = 0.

My question is, how do you know whether you will lose a root when dividing
by a function in an equation, or will it always be fairly obvious that a
solution has been lost in doing so?

Likewise, can you mistakenly introduce a solution into an equation by
multiplying by a function? For example, multiplying cosec(t) + cot(t) = 0
by (cosec(t) - cot(t)) introduces a new root. Why does this happen? Are
there certain situations where you can multiply through in an equation
without introducing a new root?

Date: 06/10/2012 at 23:33:41
From: Doctor Peterson
Subject: Re: Losing or introducing roots in a (trigonometric) equation

Hi, Hester.

In your first example, you are proving an identity. When you divide by
cos(t), you have to take note of the fact that what you did is invalid
when cos(t) is 0, so what you prove will not necessarily be true then. But
when you look at the final result, you see that sec(t) and tan(t) will,
themselves, be undefined in those cases anyway; so it is not necessary to
explicitly point out that it is not true for all t. This is probably
common in such identities.

In your second example, you are solving an equation, which is a somewhat
different situation. In the end, you just have to make sure you have no
extraneous solutions, and have not missed any. It's possible to divide as
you did, as long as you, again, stay aware that what you've done does not
cover the case where cos(t) = 0. You have to check that separately.

A wiser method is to avoid the division and just use the factored form, as
you do in solving a quadratic equation by factoring. Taking your second
example, if ...

   cos(t)(2sin(t) - 1) = 0

... then either

   cos(t) = 0      OR      2sin(t) - 1 = 0

You must solve BOTH of these, and get all the solutions. But again, I'd
never do it by division unless I saw no way around it.

So in answer to your first question: yes, extraneous solutions are common,
and whenever you multiply by an expression containing the variable, you
have to check the case where that expression is zero separately to see if
it is really valid.

To address your second question, and its example: presumably, you did 

                        cosec(t) + cot(t) = 0
   (cosec(t) + cot(t))(cosec(t) - cot(t)) = 0(cosec(t) - cot(t))
                    cosec^2(t) - cot^2(t) = 0
                               cosec^2(t) = cot^2(t)

Then maybe you multiplied both sides by sin^2(t):

                    cosec^2(t) * sin^2(t) = cot^2(t) * sin^2(t)
                                        1 = cos^2(t)
                                   cos(t) = +/-1
                                        t = pi/2 + kpi

The better way would be just to multiply by sin(t) from the start:

                        cosec(t) + cot(t) = 0
      cosec(t) * sin(t) + cot(t) * sin(t) = 0
                               1 + cos(t) = 0
                                   cos(t) = -1
                                        t = -pi/2 + 2kpi

This got only half the solutions of the longer method, missing some
extraneous solutions.

Either way, multiplying by sin(t) produced an equation that would not be
equivalent to the original if sin(t) = 0, so I have to check whether this
is true for any of my solutions -- and in fact it is ... for all of them!
I have to check whether the original equation works in this case. I find
that the cosecant and the cotangent are both undefined -- so in fact,
there are no solutions.

I'd rather avoid those multiplications entirely, so my usual method would
be to write everything in terms of sine and cosine:

                 cosec(t) +    cot(t)     = 0
                 1/sin(t) + cos(t)/sin(t) = 0

Then I'd combine the fractions to get

                       (1 + cos(t)/sin(t) = 0

This is 0 when cos(t) = -1. As above, I'd find what t has to be, but
before I even finished that, I'd have noticed that it would make 
sin(t) = 0. That would actually give us 0/0 on the left, which is

(Incidentally, you'll find something interesting if you try solving this
by graphing!)

Does that help clarify what situations mess up your solution, and what to
do when they happen (or how to avoid them)?

- Doctor Peterson, The Math Forum 

Date: 06/12/2012 at 13:48:06
From: Hester
Subject: Thank you (Losing or introducing roots in a (trigonometric) equation)

Thank you for taking the time to make such a clear explanation. It has
helped me to resolve the difficulties I was having.

Sorry to have given you this unsolvable example!

   cosec(t) + cot(t) = 0

Date: 06/12/2012 at 13:58:05
From: Doctor Peterson
Subject: Re: Thank you (Losing or introducing roots in a (trigonometric) equation)

Hi, Hester.

Right -- that made it all the more interesting to talk about!

- Doctor Peterson, The Math Forum 
Associated Topics:
High School Trigonometry

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