Rules to eliminate erroneous solutions in Trig equations?

Question

This is a High School Trig problem asking for the solution to an otherwise simple equation.

$\frac{\left(1+\cos x\right)}{\sin x}$=-1 (per comment - the domain was specified as greater than -180 degrees, less than or equal to 180 degrees.)

In order to simplify to a single function, I manipulate and square both sides

$\left(1+\cos x\right)=-\sin x$

$\cos ^2x+2\cos x+1=\sin ^2x$

Then by substituting $1-\cos ^2x$ for $\sin ^2x$ I have

$2\cos ^2x+2\cos x=0$

and using zero product rule

$\left(2\cos x\right)\left(\cos x+1\right)=0$

The first term giving $-90^{\circ}$,$90^{\circ}$ and the second $180^{\circ}$.

In hindsight, I know that when we multiply to remove the denominator (cross-multiplying) that a potential division by zero is lost. We need to review the results and invalidate the 180 degrees for this reason.

The $-90^{\circ}$ is fine, the correct answer, as cos is 0 sin is -1 and the equation fits. But, the second result, $90^{\circ}$ is also wrong. My explanation is that by using squaring as a method of seeking the roots of this equation we run into the dreaded 1 = -1 paradox, and I quickly show them how the squares of these 2 numbers are equal, but they are not.

My explanation for the 2 erroneous solutions was fine for 3 of the 7 students who asked me to explain this. The other 4 wanted a better understanding of when we can just accept the results, note that this was #20 of 30 problems, and the first one that resulted in this issue. It was also the first that offered a fraction like this, prior question were solved by factoring.

My question now asks - for this level of math are the two issues I described the only two that result in such erroneous solutions? And so the two rules -

• Check if the results cause a 'division by zero' in original equation
• Check if you've created a result from the "1 = -1 paradox"

(I think I've articulated the question fairly well. Comments or edits are welcome. Unless they are "why are they solving such problems?" I don't have a say in the material. The choice of degrees vs radians wasn't mine either.)

Answer 1

This isn't specific to trigonometry at all; it's really about understanding what it means to "solve an equation". Given functions $f, g: \mathbb{R} \to \mathbb{R}$, to solve the equation $f(x) = g(x)$ means to determine exactly which numbers $x$ make the equation true, i.e., to compute the set $\{x \in \mathbb{R}: f(x) = g(x)\}$.

There are two steps to solving an equation:

1. Assume $x$ is a number such that $f(x) = g(x)$. Under this assumption, deduce that $x$ must be an element of some set $S$ (which we want to be as small as possible to make step 2 easier).
2. For each element $x \in S$, check whether $f(x)$ is actually equal to $g(x)$.

After step 1, we have a set $S$ that contains all the solutions, but it could also contain some things that aren't solutions. Step 2 is always necessary, because everything in step 1 is done under the assumption that we're already working with a solution to the equation.

Now, it's possible to do both steps simultaneously using biconditionals — for example, if $h: \mathbb{R}^2 \to \mathbb{R}$ is a function such that $y \mapsto h(x, y)$ is injective for each $x \in \mathbb{R}$, then $f(x) = g(x)$ if and only if $h(x, f(x)) = h(x, g(x))$. (This makes precise the idea of "doing the same invertible operation to both sides". Of course, one would present this in less concise language to a high school class, but the idea is the same.) Many common student errors amount to failing to check injectivity, e.g., squaring isn't injective (although it is if we restrict to non-negative reals).

So, there are two main options: either stick to using only invertible operations (and check that all operations used really are invertible), or do step 2 explicitly by checking each possible solution. In either case, the key is to use reasoning, which requires understanding what it means to "solve an equation". A list of rules will just give the students another arbitrary thing to memorize and won't address the real issue, which is a lack of reasoning in what they're doing.

Or, to address their question directly: When can we just accept the results? Never! One way or another, we must always show the results are correct.

Answer 2

The basic issue here is that there are two kinds of manipulations on equations: reversible ones and non-reversible ones. Really, when students write out a sequence of equations, there's an implicit logical argument connecting the equations - "if this equation holds then the next equation does", and the reversible steps are bi-implications while the non-reversible ones one one directional implications.

Your students need to know the difference: squaring both sides is a non-reversible step, so it can create extra solutions. Multiplying by a number-which-might-be-0 is a non-reversible step. Most other steps they're using are reversible. If, at any point, they use a non-reversible step, they need to check the answer at the end.

This isn't the most intellectually rigorous approach, but the entire framework of manipulating equations isn't rigorous either. The underlying mathematics is that they should be making arguments, not manipulating equations, but getting them to do that conflicts with a curriculum that also wants them to solve a lot of problems in a short amount of time.

Answer 3

An alternative approach would be to solve the equation $1 + \cos{x} = -\sin{x}$ without squaring. This avoids completely the introduction of a spurious solution. One solution, that works directly with the geometric meaning of the sine and cosine, goes as follows:

1. If $\cos{x}$ is positive, then $1 + \cos{x}$ is bigger than $1$, so cannot equal the sine of any number. Thus $\cos{x}$ must be nonpositive, so $x$ must lie in the closed left half plane.
2. Since $1 + \cos{x}$ must be nonnegative, $\sin{x}$ must be nonpositive, so $x$ must lie in the closed lower half plane.
3. The preceeding remarks imply that $x$ lies in the closure of the lower left quadrant. That is $x \in [\pi, 3\pi/2]$.
4. For any angle $t$, the triangle inequality implies that $|\cos{t}| + |\sin{t}| \geq 1$ with strict inequality except when $|\cos{t}|$ and $|\sin{t}|$ are the side lengths of a degenerate triangle, that is when one of them equals $1$ and the other equals $0$. (This all can be explained geometrically, referring to a right triangle with hypotenuse $1$.) Since $1 = -\cos{x} - \sin{x} \leq |\cos{x}| + |\sin{x}|$, this implies $x$ is an integer multiple of $\pi/2$.
5. The only possible $x$ satisfying 3. and 4. and in $[-\pi, \pi]$ are $\pi$ and $-\pi/2$.

After having deduced the solution this way, one could then show the trick of squaring and observe that it can be used, provided one is careful to note that it possibly introduces false solutions.

The explanation above should work in a classroom setting. On the other hand, students probably will not be able to reproduce it on their own without substantial practice on similar problems, and so students wanting step by step recipes or thoughtless algorithms might find it frustrating.