Why cant I prove this trigonometric equation straight down?

Question

The question is as follows:

Given that $\tan^2a - 2 \tan^2b = 1$. Show that $\cos2a + \sin^2b = 0$.

After a few attempts, I successfully came up with a solution as follows:

$$\tan^2a - 2 \tan^2b = 1$$ $$(\sec^2a - 1) - 2(\sec^2b - 1) = 1$$ $$\sec^2a - 2sec^2b = 0$$ $$\frac{1}{\cos^2a} - \frac{2}{\cos^2b} = 0$$ $$\cos^2b - 2\cos^2a= 0$$ $$(1-\sin^2b) - (1 - \sin^2a) - \cos^2a = 0$$ $$\sin^2a - \cos^2a + \sin^2b = 0$$ $$\cos2a + \sin^2b = 0$$

And the proof is finished. However, before I found this solution, I came up with something like this:

$$\tan^2a - 2 \tan^2b = 1$$ $$\frac{\sin^2a}{\cos^2a} - 2\frac{\sin^2b}{\cos^2b} = 1$$ $$\sin^2a\cos^2b - 2\sin^2b\cos^2a - \cos^2a\cos^2b = 0$$

I wanted to extract the $\cos2a $ here. So I do the following

$$(\sin^2a - \cos^2a) \cos^2b - 2\sin^2b\cos^2a = 0$$ $$\cos2a + \frac{2\sin^2b\cos^2a}{\cos^2b} = 0$$

It suffices to show that

$$\frac{2\sin^2b\cos^2a}{\cos^2b} = \sin^2b$$

which is equalvalent as saying

$$\frac{\cos^2a}{\cos^2b} = \frac{1}{2}$$

Therefore, I looked at the original equation and proved that it is true like this:

$$\tan^2a - 2 \tan^2b = 1$$ $$(\sec^2a - 1) - 2(\sec^2b - 1) = 1$$ $$\sec^2a = 2sec^2b$$ $$\frac{\cos^2a}{\cos^2b} = \frac{1}{2}$$

My confusion is the reason why I need to look at the orginal equation again to finish the proof. $\tan^2a - 2 \tan^2b = 1$ is identical to $\cos2a + \sin^2b = 0$. And I didn't lose information during my proof. Shouldn't I be able to finish the proof directly like my first proof?

It may sounds like a strange question. But I hope that someone can explain it to me because it sometimes happens and is confusing to me.

Answer 1

I reckon this can be explained using ideas from algebraic geometry (of course, that specific theory only applies to polynomials, but you get the idea).

When you prove some identity from some other condition, what you are really doing is to prove the ideal generated by the identity completely contains the ideal generated by the condition. So when you "use" the equation $\tan^2 a-2\tan^2b=1$, you are really just using the fact that the function $\tan^2 a-2\tan^2 b$ is constant $1$ when restricted to the set $\{(a,b):\tan^2a-2\tan^2b=1\}$.

However, that function is not constant outside of that set, so it changes the "shape" of the ideal outside. What we want to do is to fine-tune the shape so that it coincides with the condition. If you use the equation in the wrong way, although the "shape" is changed, but it's not in the correct direction. So we have to use that equation again to get it right.

Hope this gives some intuition for you :).

Answer 2

Starting from this line continue as follows (instead of dividing by $\cos ^2b$, express it as $1-\sin^2b$): $$(\sin^2a - \cos^2a) \cos^2b - 2\sin^2b\cos^2a = 0 \\ (\sin^2a - \cos^2a) (1-\sin^2b) - 2\sin^2b\cos^2a = 0\\ -\cos 2a-\sin^2a\sin^2b+\sin^2b\cos^2a-2\sin^2b\cos^2a=0 \\ -\cos2a-\sin^2a\sin^2b-\sin^2b\cos^2a=0\\ \cos2a+\sin^2b=0.$$

Answer 3

Why are you not comfortable about the structure of your first proof? With the little claim about the case when $\sin b=0$ mentioned in the comments under OP, it's OK.

What you did is that you managed to reduce the problem to showing that the equality $$\frac{\cos^2a}{\cos^2b}=\frac 12$$ is true. Then you used the given hypothesis to prove this (more manageable equation). There's nothing unpleasant at all about this. It's very valid. That you reduced the original equation to the one above doesn't mean you can no longer use the original, or that it has disappeared. You may still use it, and that's right.

You're not very specific about why you're worried about this approach, but if you think it's circular then that's not right. You want to show that an equation implies another. You then show that this implication is true if a certain equation holds. Then you prove that this equation holds by using the original equation (which you can use as many times as you wish in the problem -- there's no constraint that says you must use the hypothesis only once). This is not circular, and indeed I like that approach better; what's happening here is something like a composition of hypotheses, as it were -- You want to show that $A$ implies $C.$ You then prove that if $A,$ and if $A \implies B,$ then $C;$ you then show that $A$ implies $B.$ Then your result followed. And that's fine -- it's only that the structure is a bit complicated, nothing more.

I hope this clears the air for you. :)