Methodology for Verifying Trigonometric Identities

Question

Lets say I have an equation like: $$\frac{\sin^2x+\cos^2x}{\cos^2x\sec^2x}=1$$

Our teacher said you have to verify the equality by simplifying the left hand side or the right hand side without moving(doing operations to both sides) anything from one side of the equation to the other. I have a problem with this because I proposed that we can verify the equality by multiplying the denominator by 1 and moving it to the other side. Then you can proceed to verify the equation by substituting trig identities. My argument is that if correct algebraic manipulation is done, then the equality is the same. It won't suddenly become false or true.

My method: **the addition to the classical method is that you can multiply the equation by both sides while preserving the equality which is specifically not multiplying by 0 or applying operators such as derivatives or squaring etc as these do not preserve equalities.

$$\sin^2x+\cos^2x = \cos^2x·\sec^2x$$ $$1=\frac{\cos^2x·1}{\cos^2x}$$ $$1=1$$ equality is true

It requires ingenuity to be able to do the right manipulations...

Is my teacher justified in requiring that nothing is moved between the sides? is there a convincing argument for my side or am I wrong? I would be grateful if someone can enlighten me; just why is it wrong to do algebraic manipulation before verifying an equality? Keep in mind that this method is meant to be extended to the more complicated problems. I just gave a simple example to illustrate it.

For reiteration and clarity, my question is why is my method wrong? What evidence is available to back up any certain answer?

Answer 1

Consider the equation $1 = -1$. Allow me to prove this equation by performing the same operation on both sides, namely squaring. Hence, $1^2 = (-1)^2$ or $1 =1 $. Therefore, the first equation must be true. Does this seem like sound logic? The key observation here is that a false statement can be turned into a true statement if you abuse properties that are true for true statements. "Do the same thing" to equal values does indeed preserve their equality, but "doing the same thing" to unequal values can make them "equal".

You may find this related queation enlightening on on the logic consequences of how you proceed with problems such as this.

Answer 2

I think your equality should be $\displaystyle \frac{\sin^2x+\cos^2x}{\cos^2x\sec^2x}=1$.

When I read your work, I naturally interpret it as

$\displaystyle \frac{\sin^2x+\cos^2x}{\cos^2x\sec^2x}=1 \implies \sin^2x+\cos^2x=\cos^2x\sec^2x \implies 1=\frac{\cos^2x\cdot1}{\cos^2x} \implies 1=1$.

It does not make sense. I already know that $1=1$, but it is not because of $\displaystyle \frac{\sin^2x+\cos^2x}{\cos^2x\sec^2x}=1$. $\displaystyle \frac{\sin^2x+\cos^2x}{\cos^2x\sec^2x}=1$ is what you want to show and you should not start from it.

The correct sequence should be

$\displaystyle 1=1 \implies1=\frac{\cos^2x\cdot1}{\cos^2x} \implies \sin^2x+\cos^2x=\cos^2x\sec^2x \implies\frac{\sin^2x+\cos^2x}{\cos^2x\sec^2x}=1$.

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Let me 'prove' that $ \sin^2x+ \cos^2 x=2$.

$ \sin^2x+ \cos^2 x=2 \implies (0)(\sin^2x+ \cos^2 x)=(0)(2) \implies 0=0$

equality is true.

Can you see the danger of the approach?

Answer 3

As you observe, multiplying by zero is the only unreversable step. (Well...that's not quite true, but close enough.)

The problem with applying this in the context of proving trig identities is that it's sometimes tough to know that you are multiplying by zero.

For instance, you might multiply both sides by $2\sin^4 x + \sin^2 (2x) + 2 \cos^4 x - 1$. Are you multiplying by zero or not? As it turns out, you're multiplying by zero in this case ... but you don't know that until you've proved another trig identity, namely that $$ 2\sin^4 x + \sin^2 (2x) + 2 \cos^4 x - 1 = 0. $$ You can see how this can lead you down a rathole pretty fast.

So your logic (with appropriate restructuring) is fine: if you apply an invertible operations to both sides of your supposed equality, and the resulting equality is true, then so was your supposed equality.

But in practice, it's almost useless, because especially in the context of trig identities, it's tough to know whether a given function is injective or not. In particular, multiplication by some trigonometric expression may turn out to be multiplication by zero, and then it's "game over."

Let me give one last example for you to consider: $$ \frac{\sin (2x)}{\sin x} = 2\cos x $$ If we multiply both sides by $\sin x$, we get $$ \sin (2x) = 2 \sin x \cos x $$ which is true. But the original statement is false. Why? Because at $x = 0$, the left hand side is undefined, while the right-hand side is $2 \cos 0 = 2$.

Notice that in this example, I did an apparently reversible operation (multiplying by $\sin x$, which is certainly not the 0 function!) and derived a true statement, but the original statement was not true.

When you can carefully explain this example, and how you might avoid situations like this one, you can probably safely start ignoring your teacher's advice.

Answer 4

As I think, you have obviously removed $\cos^2 x\sec^2 x$ from LHS to RHS (i.e. cross-manipulated the both sides) which may make you teacher object to you! A better and more standard way (to fit your teacher's will) is that simplify $\cos^2x\sec^2x$ and substitute it with $1$ (except when $\cos x=0$ since ambiguity happens) and get onto an always-true statement $\sin^2x+\cos^2x=1$. Therefore the answer is $$\Bbb R-\left\{k\pi+{\pi\over 2}\ |\ k\in\Bbb Z\right\}$$