How do I discover a hidden Greatest Common Factor in a trigonometric rational function?

A

Al-Layth

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The general method I execute in trying to verify trigonometric identities is to
1.) rewrite everything in terms of sin and cos
2.) sum the terms together, i do this for both sides of the identity

3.) now at this point if the identity hasn't been verified, the problem arises where both sides contain a single term, but one side's term is more complicated than the other.
Sometimes a greatest common factor is staring me in the face, here it can be easy to cancel down into the simpler side thus the identity is verified.

However, The problem I am concerned with is when the more complicated side's numerator and denominator do not contain a Greatest common factor, as they are right now. Simple example of this( i say simple but i spent a solid ten minutes on it lol )

[math]\frac{1-\sin^4(x)}{\cos^4(x)}=\frac{1+\sin^2(x)}{\cos^2(x)}[/math]
As we can see there is one term on both sides of the identity, yet one term is more complicated. this is the one i must reduce to the other.
DoTs works here;

[math]\frac{(1-\sin^2(x))(1+\sin^2(x))}{\cos^2(x)(1-\sin^2(x))}=\frac{1+\sin^2(x)}{\cos^2(x)}[/math]




So in the pursuit of verifying a trig identity ( assuming the identity is actually true) with this specific strategy, eventually

it becomes a game of manipulating the numerator and denominator with the available body of trig identities until a GCF can be extracted.​

I am looking for a general method of doing that. Or is it all done by "experience" and "intuition" and "doing lots of problems"?

Now my progress in this problem is as follows:
- since i ensured i wrote everything in the identity in terms of sin and cos, i can ignore any identity that contains a trig term other than sin or cos or their natural powers. this leaves very few identities which is good.

- Analysing the differences in the respective numerators and respective denominators.

sign changes suggest the pythagorean identity must be used
differences of powers of exactly two(or perhaps any multiple of two) suggest the pythagorean identity is required as well.

- difference of two squares rule has also been helpful

any advice? thx
 
Al-Layth said:
The general method I execute in trying to verify trigonometric identities is to
1.) rewrite everything in terms of sin and cos
2.) sum the terms together, i do this for both sides of the identity

3.) now at this point if the identity hasn't been verified, the problem arises where both sides contain a single term, but one side's term is more complicated than the other.
Sometimes a greatest common factor is staring me in the face, here it can be easy to cancel down into the simpler side thus the identity is verified.

However, The problem I am concerned with is when the more complicated side's numerator and denominator do not contain a Greatest common factor, as they are right now. Simple example of this( i say simple but i spent a solid ten minutes on it lol )

[math]\frac{1-\sin^4(x)}{\cos^4(x)}=\frac{1+\sin^2(x)}{\cos^2(x)}[/math]
As we can see there is one term on both sides of the identity, yet one term is more complicated. this is the one i must reduce to the other.
DoTs works here;

[math]\frac{(1-\sin^2(x))(1+\sin^2(x))}{\cos^2(x)(1-\sin^2(x))}=\frac{1+\sin^2(x)}{\cos^2(x)}[/math]




So in the pursuit of verifying a trig identity ( assuming the identity is actually true) with this specific strategy, eventually

it becomes a game of manipulating the numerator and denominator with the available body of trig identities until a GCF can be extracted.​

I am looking for a general method of doing that. Or is it all done by "experience" and "intuition" and "doing lots of problems"?

Now my progress in this problem is as follows:
- since i ensured i wrote everything in the identity in terms of sin and cos, i can ignore any identity that contains a trig term other than sin or cos or their natural powers. this leaves very few identities which is good.

- Analysing the differences in the respective numerators and respective denominators.

sign changes suggest the pythagorean identity must be used
differences of powers of exactly two(or perhaps any multiple of two) suggest the pythagorean identity is required as well.

- difference of two squares rule has also been helpful

any advice? thx
Click to expand...

Do not forget the rules of factoring difference of cubes. (or summation of cubes).
 
Al-Layth said:
I am looking for a general method of doing that. Or is it all done by "experience" and "intuition" and "doing lots of problems"?
Click to expand...
I have no interest in trying to find a universal method; there is too much variation in these problems. I would describe "experience and intuition" as "being familiar with many heuristics", which include things like the following.

Like you, I commonly (but not always) use sines and cosines, to reduce the number of identities I need to consider; and I rewrite everything in terms of a single angle (using cofunction or double angle identities, for example). Then I work on the more complicated side, looking for anything I can do to make it change in the direction of the other side. (For example, if the right side is a single term, I want to make a single term on the left. If the right side has only cosines, I might want to get only cosines.) If the other side is also somewhat complicated, I will take a few steps to simplify that first, to give me a simpler target,

Then, I find that it is often a matter of alternating trig identities with algebraic work such as adding fractions when the goal is a single term, or splitting a fraction when the goal is two terms, or factoring in order to cancel. For the trig steps, a squared function (more than a sign) suggests the Pythagorean identity. But there are other tricks too; one is to multiply numerator and denominator of a fraction by a sort of conjugate, which will result in some squared functions.

In the case you are asking about,
Al-Layth said:
[math]\frac{1-\sin^4(x)}{\cos^4(x)}=\frac{1+\sin^2(x)}{\cos^2(x)}[/math]
Click to expand...
I would quickly see the difference of squares and factor it, not because I already see that there will be a common factor, but just because it seems likely to open up the problem and reveal something. Then I see [imath]1-\sin^2(x)[/imath] and that it can be written as [imath]\cos^2(x)[/imath], which will cancel with the denominator. When you're lost in the woods, sometimes you just take a step to see if anything new becomes visible from a new location. That's what we've done here.

In general, in any kind of proof, you just have to try anything you see that might work, and persevere. Don't expect the first thing you try to give the answer; but after "doing lots of problems" and making a list of methods that have worked, you will have more things to try! This is the nature of problem solving in general; see Polya, who provides many general heuristics.
 
Dr.Peterson said:
I have no interest in trying to find a universal method; there is too much variation in these problems. I would describe "experience and intuition" as "being familiar with many heuristics", which include things like the following.

Like you, I commonly (but not always) use sines and cosines, to reduce the number of identities I need to consider; and I rewrite everything in terms of a single angle (using cofunction or double angle identities, for example). Then I work on the more complicated side, looking for anything I can do to make it change in the direction of the other side. (For example, if the right side is a single term, I want to make a single term on the left. If the right side has only cosines, I might want to get only cosines.) If the other side is also somewhat complicated, I will take a few steps to simplify that first, to give me a simpler target,

Then, I find that it is often a matter of alternating trig identities with algebraic work such as adding fractions when the goal is a single term, or splitting a fraction when the goal is two terms, or factoring in order to cancel. For the trig steps, a squared function (more than a sign) suggests the Pythagorean identity. But there are other tricks too; one is to multiply numerator and denominator of a fraction by a sort of conjugate, which will result in some squared functions.

In the case you are asking about,

I would quickly see the difference of squares and factor it, not because I already see that there will be a common factor, but just because it seems likely to open up the problem and reveal something. Then I see [imath]1-\sin^2(x)[/imath] and that it can be written as [imath]\cos^2(x)[/imath], which will cancel with the denominator. When you're lost in the woods, sometimes you just take a step to see if anything new becomes visible from a new location. That's what we've done here.

In general, in any kind of proof, you just have to try anything you see that might work, and persevere. Don't expect the first thing you try to give the answer; but after "doing lots of problems" and making a list of methods that have worked, you will have more things to try! This is the nature of problem solving in general; see Polya, who provides many general heuristics.
Click to expand...
thanks for the advice

Wouldn't converting the sines and cosines into their complex exponential forms be a perfectly general technique??

if the identity is true, then both sides should easily simplify into each other. I have not done it on many problems yet though.
 
Al-Layth said:
thanks for the advice

Wouldn't converting the sines and cosines into their complex exponential forms be a perfectly general technique??

if the identity is true, then both sides should easily simplify into each other. I have not done it on many problems yet though.
Click to expand...
Interesting thought. I'll be interested to see your best and worst examples. It can imagine a couple things that might go wrong, but it may work just fine.
 
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