Are there multiple ways to verify a trig identity?

[ProfuselyQuarky]

Well, are there? I thought that problems involving the verification of identities pretty much checked themselves because you know whether the steps you’re doing are legitimate or not and, of course, you know whether you’ve reached the expression you want. However, I got one of these problems wrong and I don’t why. So, if there are multiple ways to verify trig identities, are some ways better or “more correct” than others?

 

[BvU]

There are many ways to Rome (there used to be a time all ways led to Rome).

However, I got one of these problems wrong and I don’t why.

Wouldn't it be a good road out of frustration to post the thing and show what you did ? Perhaps someone can help you see the way out.
The current answer to your outcry would be something like: yes, but don't ask me to prove it.

So, if there are multiple ways to verify trig identities, are some ways better or “more correct” than others?

Only for some teachers. In reality correct is correct and vice versa..

 

[jedishrfu]

There are multiple ways to verify them depending which identity you start with.

 

[fresh_42]

Only for some teachers. In reality correct is correct and vice versa..

Oh, how I love this line! If I didn't know it were a misuse of expression I would reply: not some, almost all!

 

[BvU]

Hehe,
I'm not an expert in proof theory, but I know for sure that if you prove something correctly in different ways, they all count as proof. 'Correct' means flawless, zero mistakes. More correct than that isn't posiible; equally correct is. Some folks find an elegant proof preferable, which is fine.

 

[ProfuselyQuarky]

Oh, yeah, sorry, about that. Anyway, here is just one of them:

##\frac {\cos^2\theta-1}{\cos^2\theta}=-\sec^2\theta \sin^2\theta ##

Really simple, right?

##\frac {\cos^2\theta-1}{\cos^2\theta}=\sec^2\theta(-\sin^2\theta)##

##\frac {\cos^2\theta-1}{\cos^2\theta}=\frac {-\sin^2\theta}{\cos^2 \theta}##

##\frac {\cos^2\theta-1}{\cos^2\theta}=\frac {-(1-\cos^2\theta)}{\cos^2 \theta}##

##\frac {\cos^2\theta-1}{\cos^2\theta}=\frac {\cos^2\theta-1}{\cos^2 \theta}##

 

[cnh1995]

I got one of these problems wrong and I don’t why.

Been there! I argued with my professor over a trig proof, for which he thought my method was wrong. Actually ,the method was not in his lecture-notes, so I got it wrong. After a heated argument with him, I ended up out of the class for the rest of the lecture! I never did that again...

 

[cnh1995]

Oh, yeah, sorry, about that. Anyway, here is just one of them:

##\frac {\cos^2\theta-1}{\cos^2\theta}=-\sec^2\theta \sin^2\theta ##

Really simple, right?

##\frac {\cos^2\theta-1}{\cos^2\theta}=\sec^2\theta(-\sin^2\theta)##

##\frac {\cos^2\theta-1}{\cos^2\theta}=\frac {-\sin^2\theta}{\cos^2 \theta}##

##\frac {\cos^2\theta-1}{\cos^2\theta}=\frac {-(1-\cos^2\theta)}{\cos^2 \theta}##

##\frac {\cos^2\theta-1}{\cos^2\theta}=\frac {\cos^2\theta-1}{\cos^2 \theta}##

You proved RHS=LHS. Perhaps your teacher wanted you to start with the LHS and prove LHS=RHS.

 

[ProfuselyQuarky]

Been there! I argued with my professor over a trig proof, for which he thought my method was wrong. Actually ,the method was not in his lecture-notes, so I got it wrong. After a heated argument with him, I ended up out of the class for the rest of the lecture! I never did that again...

It happens—usually when I am in a dispute, I get my way (you’ve got to know how to talk or email the right way) but from what it sounds like, you’re talking about a college-level class, so I don’t know. That sounds bad.

You proved RHS=LHS. Perhaps your teacher wanted you to start with the LHS and prove LHS=RHS.

I wrote the first equation the wrong way, just a typo. But, as long as I show that the two expressions are equivalent, that shouldn’t matter. For longest time, I’ve been taught to deal with the easier side . . .

 

[Samy_A]

Oh, yeah, sorry, about that. Anyway, here is just one of them:

##\frac {\cos^2\theta-1}{\cos^2\theta}=-\sec^2\theta \sin^2\theta ##

Really simple, right?

##\frac {\cos^2\theta-1}{\cos^2\theta}=\sec^2\theta(-\sin^2\theta)##

##\frac {\cos^2\theta-1}{\cos^2\theta}=\frac {-\sin^2\theta}{\cos^2 \theta}##

##\frac {\cos^2\theta-1}{\cos^2\theta}=\frac {-(1-\cos^2\theta)}{\cos^2 \theta}##

##\frac {\cos^2\theta-1}{\cos^2\theta}=\frac {\cos^2\theta-1}{\cos^2 \theta}##

Maybe he didn't like the way you wrote this down.
If you have to prove that ##\frac {\cos^2\theta-1}{\cos^2\theta}=-\sec^2\theta \sin^2\theta ##, it may be clearer to write it like this (using your steps):
##\displaystyle -\sec^2\theta \sin^2\theta = \sec^2\theta(-\sin^2\theta)=\frac {-\sin^2\theta}{\cos^2 \theta}=\frac {-(1-\cos^2\theta)}{\cos^2 \theta}=\frac {\cos^2\theta-1}{\cos^2 \theta}## QED.

 

[cnh1995]

. But, as long as I show that the two expressions are equivalent, that shouldn’t matter

Exactly!

 

[ProfuselyQuarky]

Maybe he didn't like the way you wrote this down.
If you have to prove that ##\frac {\cos^2\theta-1}{\cos^2\theta}=-\sec^2\theta \sin^2\theta ##, it may be clearer to write it like this (using your steps):
##\displaystyle -\sec^2\theta \sin^2\theta = \sec^2\theta(-\sin^2\theta)=\frac {-\sin^2\theta}{\cos^2 \theta}=\frac {-(1-\cos^2\theta)}{\cos^2 \theta}=\frac {\cos^2\theta-1}{\cos^2 \theta}## QED.

Perhaps, but I don't think so. For every problem and proof I have had to complete, I wrote it out the same way that I just did (including other identities) and no one has complained. So from what I’m reading nothing is exactly “wrong”?

 

[ProfuselyQuarky]

The problem I posted is just one of the easy examples. I haven't even bothered posting the ones that take a whole page or so . . .

 

[robphy]

Can you post the actual statement of the question?
(Are there certain requirements that are specified?)

 

[ProfuselyQuarky]

Can you post the actual statement of the question?
(Are there certain requirements that are specified?)

Sure. As follows:

Verify each identity.

I'm not kidding.

 

[Samy_A]

Perhaps, but I don't think so. For every problem and proof I have had to complete, I wrote out the same way that I just did (including other identities) and no one has complained. So from what I’m reading nothing is exactly “wrong”?

Good for you that I'm not your teacher.

Yes, none of the 4 equations is wrong, but we don't know that until the last. Maybe this specific example is so simple that we actually do see immediately that they are all correct.
But for a more difficult identity (say ##\sin^4x +\cos^4 x =\frac{3}{4}+\frac{1}{4}\cos{4x}##), I probably would object to your way of writing the solution.

 

[ProfuselyQuarky]

But for a more difficult identity (say sin4x+cos4x=34+14cos4x\sin^4x +\cos^4 x =\frac{3}{4}+\frac{1}{4}\cos{4x}), I probably would object to this way of writing the solution.

Oh, well, I should do that from now on then.

But, that’s also why I’m stumped. This specific identity is SO simple . . .

 

[zinq]

I have to agree – when you prove something, you not only want to make sure that each step follows from the previous ones, but you also want to make sure that the last line is that which was to be proved. In fact, that is the definition of a proof.

Normally if I'm asked to verify an identify of the form LHS = RHS, I will start with the LHS and manipulate it from one formula to the next so that each one is equal to the last. If that isn't convenient or doesn't suffice, you could also do the same with the RHS — until the manipulated LHS and the manipulated RHS are identical.

For the identity in question, it's really just a matter of knowing that

1/cosθ = secθ​

and that

cos²θ + sin²θ = 1.​

 

[micromass]

You can easily dodge the criticism in this thread by using the ##\Leftrightarrow## symbol. So:

##\frac {\cos^2\theta-1}{\cos^2\theta}=\sec^2\theta(-\sin^2\theta)##

##\Leftrightarrow~~\frac {\cos^2\theta-1}{\cos^2\theta}=\frac {-\sin^2\theta}{\cos^2 \theta}##

##\Leftrightarrow~~\frac {\cos^2\theta-1}{\cos^2\theta}=\frac {-(1-\cos^2\theta)}{\cos^2 \theta}##

##\Leftrightarrow~~\frac {\cos^2\theta-1}{\cos^2\theta}=\frac {\cos^2\theta-1}{\cos^2 \theta}##

 

[ProfuselyQuarky]

You can easily dodge the criticism in this thread by using the ##\Leftrightarrow## symbol. So:

##\frac {\cos^2\theta-1}{\cos^2\theta}=\sec^2\theta(-\sin^2\theta)##

##\Leftrightarrow~~\frac {\cos^2\theta-1}{\cos^2\theta}=\frac {-\sin^2\theta}{\cos^2 \theta}##

##\Leftrightarrow~~\frac {\cos^2\theta-1}{\cos^2\theta}=\frac {-(1-\cos^2\theta)}{\cos^2 \theta}##

##\Leftrightarrow~~\frac {\cos^2\theta-1}{\cos^2\theta}=\frac {\cos^2\theta-1}{\cos^2 \theta}##

Sure, I'll use that symbol. It looks rather cool, but why is my answer "wrong"??

 

[micromass]

Sure, I'll use the "therefore" symbol. It looks rather cool, but why is my answer "wrong"??

Nono, it's not a therefore symbol. This is where many newbies go wrong. If you want to prove something, then using ##\Rightarrow## is not a good idea.

For example, you could do the following to solve ##2x + 3 = 5##:

[tex]2x + 3 = 5~~\Rightarrow~~ 2x = 2~~\Rightarrow x = 1[/tex]

None of that implies that ##1## is the solution you seek. All it does is show that ##1## is the only possible solution.

Let's clarify by making an absolutely wrong proof that ##4 = 2##.

[tex]4 = 2~~\Rightarrow ~~ 0\cdot 4 = 0\cdot 2 ~~\Rightarrow~~ 0=0[/tex]

This is a bit what you had in mind, and it's obviously wrong. Going from an expression to an expression that's true does not mean the original expression is true! The only way you can be certain of this is if also every step is reversible, this is what ##\Leftrightarrow## means. So it is not a therefore symbol. The following would be incorrect ## 4 = 2 ~~\Leftrightarrow 0\dot 4 = 0\cdot 2##, while ##4 = 2~~\Rightarrow ~~ 0\cdot 4 = 0\cdot 2## or "##4= 2## therefore ##0\cdot 4 = 0\cdot 2##" is absolutely correct.

 

[ProfuselyQuarky]

Nono, it's not a therefore symbol.

Oops, yeah, sorry, that's why I edited my post

Thanks for the explanation, it makes sense.

 

[Vanadium 50]

The problem is simpler that most of the replies suggest. When proving a trig identity, you can only work one side of the equation. The reason is that the equation is what you have set out to prove, so you cannot use this "fact" until you have actually proved it.

 

[Mark44]

Oh, yeah, sorry, about that. Anyway, here is just one of them:

##\frac {\cos^2\theta-1}{\cos^2\theta}=-\sec^2\theta \sin^2\theta ##

Really simple, right?

##\frac {\cos^2\theta-1}{\cos^2\theta}=\sec^2\theta(-\sin^2\theta)##

##\frac {\cos^2\theta-1}{\cos^2\theta}=\frac {-\sin^2\theta}{\cos^2 \theta}##

##\frac {\cos^2\theta-1}{\cos^2\theta}=\frac {-(1-\cos^2\theta)}{\cos^2 \theta}##

##\frac {\cos^2\theta-1}{\cos^2\theta}=\frac {\cos^2\theta-1}{\cos^2 \theta}##

The problem is simpler that most of the replies suggest. When proving a trig identity, you can only work one side of the equation. The reason is that the equation is what you have set out to prove, so you cannot use this "fact" until you have actually proved it.

The problem with your work, @ProfuselyQuarky, which is suggested at in Vanadium's post, is that you have proved that ##\frac {\cos^2\theta-1}{\cos^2\theta}## is equal to itself (Well, duh!).

If you start with one side and show by a succession of identities that you end up with what's on the right side, you have proved the identity.

The goal is to proved that ##\frac {\cos^2\theta-1}{\cos^2\theta}=-\sec^2\theta \sin^2\theta ##, NOT that ##\frac {\cos^2\theta-1}{\cos^2\theta}=\frac {\cos^2\theta-1}{\cos^2\theta} ##. If all of the steps you applied are reversible (what is what the ##\Leftrightarrow## that micromass wrote is used for), then you can write the proof correctly, showing that the LHS is indeed equal to the RHS.

 

[member 587159]

I agree, I read that is logically incorrect to end a proof with p = p, since that would be a tautology (something that's always true)

 

[ProfuselyQuarky]

The problem is simpler that most of the replies suggest. When proving a trig identity, you can only work one side of the equation. The reason is that the equation is what you have set out to prove, so you cannot use this "fact" until you have actually proved it.

The goal is to proved that ##\frac {\cos^2\theta-1}{\cos^2\theta}=-\sec^2\theta \sin^2\theta ##, NOT that ##\frac {\cos^2\theta-1}{\cos^2\theta}=\frac {\cos^2\theta-1}{\cos^2\theta} ##. If all of the steps you applied are reversible (what is what the ##\Leftrightarrow## that micromass wrote is used for), then you can write the proof correctly, showing that the LHS is indeed equal to the RHS.

Alright, then. All I would have to do is add the ##\Leftrightarrow## and then the proof is immediately legitimate? Thanks! I'll definitely do that from now on. I wonder why I was never taught that, then . . .

Anyway, just one little question regarding ##\Leftrightarrow##. While figuring out how to type it up, I saw that there was a difference between \Leftrightarrow and \Longleftrightarrow. Is there a difference in what they're used for? Does using one or the other in the context of this thread alter the meaning in anyway? Especially when doing my work on paper, it'd probably be hard to distinguish the two if I wrote it.

 

[micromass]

Alright, then. All I would have to do is add the ##\Leftrightarrow## and then the proof is immediately legitimate? Thanks! I'll definitely do that from now on. I wonder why I was never taught that then, though . . .

Yes, but there is a danger. You can't just write ##\Leftrightarrow## between everything, you really need to be sure that you can go from left to right and right to left. Reasonable looking statements like

[tex]x=y~\Leftrightarrow 1 = \frac{x}{y}[/tex]

or

[tex]x^2 = y~\Leftrightarrow~x=\sqrt{y}[/tex]

are wrong for example. In my math courses, I highly discourage the use of ##\Leftrightarrow## for these reasons.

Anyway, just one little question regarding ##\Leftrightarrow##. While figuring out how to type it up, I saw that there was a difference between \Leftrightarrow and \Longleftrightarrow. Is there a difference in what they're used for? Does using one or the other in the context of this thread alter the meaning in anyway? Especially when doing my work on paper, it'd probably be hard to distinguish the two if I wrote it.

No, there is no difference between them.

 

[ProfuselyQuarky]

You can't just write ##\Leftrightarrow## between everything, you really need to be sure that you can go from left to right and right to left.

In my math courses, I highly discourage the use of ##\Leftrightarrow## for these reasons.

Okay, I'll just have to pay attention more when using it, which will only be when verifying identities.

Thanks everyone! I felt like this was a dumb question, but it's helped!

 

[Mark44]

Reasonable looking statements like
[tex]x=y~\Leftrightarrow 1 = \frac{x}{y}[/tex]
or
[tex]x^2 = y~\Leftrightarrow~x=\sqrt{y}[/tex]
are wrong for example. In my math courses, I highly discourage the use of ##\Leftrightarrow## for these reasons.

In both examples the solution set for the equation on the left is different from the equation on the right. In the first example, the solution set on the left consists of pairs of real numbers such that x = y. The solution set for the equation on the right is almost the same, but doesn't include (0, 0). Dividing both sides of an equation is not necessarily reversible, particularly if there's the possibility of dividing by zero.

In the second example, the solution set of the equation on the left consists of pairs of numbers (x, y) such that ##y = x^2##, including the pairs (1, 1), (-1, 1), (2, 4), (-2, 4), and so on. The solution set for the equation on the right does not include (-1, 1), (-2, 4), and so on. Taking the square root of both sides of an equation is not a reversible step.

 

[ProfuselyQuarky]

That makes plenty of sense

When I think about, it's not that difficult to determine whether a move is legit or not.

 

[Mark44]

That makes plenty of sense

When I think about, it's not that difficult to determine whether a move is legit or not.

No, it's not. If the operation you apply to both sides of an equation is one-to-one, it's reversible. For example, adding a number to or subtracting a number from each side, multiplying each side by a nonzero number, and exponentiating each side are reversible operations. In contrast, squaring both sides of an equation is not reversible.

 

[Vanadium 50]

When I think about, it's not that difficult to determine whether a move is legit or not.

If you do these by operating on only one side - the way trig has traditionally been taught - it's automatic.