# B Are there multiple ways to verify a trig identity?

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1. May 1, 2016

### 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?

2. May 1, 2016

### BvU

There are many ways to Rome (there used to be a time all ways led to Rome).
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.
Only for some teachers. In reality correct is correct and vice versa..

3. May 1, 2016

### Staff: Mentor

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

4. May 1, 2016

### Staff: Mentor

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

5. May 1, 2016

### 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.

6. May 1, 2016

### 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?

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

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

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

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

7. May 1, 2016

### cnh1995

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...

8. May 1, 2016

### cnh1995

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

9. May 1, 2016

### ProfuselyQuarky

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.
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 . . .

10. May 1, 2016

### Samy_A

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.

11. May 1, 2016

### cnh1995

Exactly!

12. May 1, 2016

### ProfuselyQuarky

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”?

Last edited: May 1, 2016
13. May 1, 2016

### 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 . . .

14. May 1, 2016

### robphy

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

15. May 1, 2016

### ProfuselyQuarky

Sure. As follows:

I'm not kidding.

16. May 1, 2016

### Samy_A

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.

17. May 1, 2016

### ProfuselyQuarky

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

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

18. May 1, 2016

### 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.​

19. May 1, 2016

### micromass

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

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

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

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

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

20. May 1, 2016

### ProfuselyQuarky

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

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