simplification problem

[allegansveritatem]

Here is expression to be simplified:


I couldn't get anywhere with this., or rather, I got this far:

(
Am I right in supposing that this expression is not available to any further simplification? (I do not consider the second image as any kind of simplification.)

 

[MarkFL]

Try factoring both numerator and denominator...

 

[Deleted member 4993]

If I were to do this problem - I would substitute:

\(\displaystyle u \ = \ \cot(\alpha) \ \)....... In my opinion this will make your expression look simpler.

 

[Steven G]

What does --> mean in your work?
Try converting everything to sines and cosines or do what Khan suggested.

 

[pka]

Converting to all sines & cosines gives a nightmare.
\(\displaystyle \dfrac{\cot^2(x)-4}{\cot^2-\cot(x)-6}=\dfrac{\cos^2(x)-4\sin^2(x)}{\cos^2(x)-\sin(x)\cos(x)-6\sin^2(x)}\)
Do you see what I mean? If so then use Mr. Khan's suggestion .

 

[Dr.Peterson]

Putting everything together, I would factor (using a substitution if it helped me see how to do that, but then returning to the original), and possibly change to sines and cosines at the end. (What is considered simplest is debatable, but may be implied by the exact wording of the instructions, which have not been given.)

 

[allegansveritatem]

If I were to do this problem - I would substitute:

\(\displaystyle u \ = \ \cot(\alpha) \ \)....... In my opinion this will make your expression look simpler.

Seems like the way to go. Thanks

 

[allegansveritatem]

What does --> mean in your work?
Try converting everything to sines and cosines or do what Khan suggested.

I use the arrow to say something like: "since the preceding is true, then it follows that ......"

I am going to try the substitution suggestion.

 

[allegansveritatem]

Converting to all sines & cosines gives a nightmare.
\(\displaystyle \dfrac{\cot^2(x)-4}{\cot^2-\cot(x)-6}=\dfrac{\cos^2(x)-4\sin^2(x)}{\cos^2(x)-\sin(x)\cos(x)-6\sin^2(x)}\)
Do you see what I mean? If so then use Mr. Khan's suggestion .

Yes, all that junk needs some method to clean it up.

 

[allegansveritatem]

Putting everything together, I would factor (using a substitution if it helped me see how to do that, but then returning to the original), and possibly change to sines and cosines at the end. (What is considered simplest is debatable, but may be implied by the exact wording of the instructions, which have not been given.)

I will try factoring and substitution. I get the feeling that the only way to get reasonably proficient at handling these trigonometric functions is the way of the musician, i.e., practice and more of same.

 

[MarkFL]

Try factoring both numerator and denominator...

[MATH]\frac{\cot^2(\alpha)-4}{\cot^2(\alpha)-\cot(\alpha)-6}=\frac{(\cot(\alpha)+2)(\cot(\alpha)-2)}{(\cot(\alpha)+2)(\cot(\alpha)-3)}=\frac{\cot(\alpha)-2}{\cot(\alpha)-3}[/MATH]

 

[allegansveritatem]

[MATH]\frac{\cot^2(\alpha)-4}{\cot^2(\alpha)-\cot(\alpha)-6}=\frac{(\cot(\alpha)+2)(\cot(\alpha)-2)}{(\cot(\alpha)+2)(\cot(\alpha)-3)}=\frac{\cot(\alpha)-2}{\cot(\alpha)-3}[/MATH]

You won't believe this but I got this exact same result today--independently--using the substitution method, thus:


So, I guess what they say is true about great minds?

 

[Dr.Peterson]

Good work.

But don't forget that "cot" by itself is meaningless in an expression -- it's a function name, not a quantity you can subtract 2 from. So your final answer must include alpha as the argument of each cot.

And, again, there are different valid answers depending on what you call simple. One that hasn't been mentioned is that you could multiply the numerator and denominator by [MATH]\tan\alpha[/MATH] if you like tan more than cot.

 

[Steven G]

You won't believe this but I got this exact same result today--independently--using the substitution method, thus:
View attachment 15762

So, I guess what they say is true about great minds?

Can you please use equal signs. The problem is if you don't then the beginning does not equal the end--or at least you can't conclude that.

[MATH]\frac{\cot^2(\alpha)-4}{\cot^2(\alpha)-\cot(\alpha)-6}=\frac{(\cot(\alpha)+2)(\cot(\alpha)-2)}{(\cot(\alpha)+2)(\cot(\alpha)-3)}=\frac{\cot(\alpha)-2}{\cot(\alpha)-3}[/MATH]

Did I really say to convert to sines and cosines? I need to use my eye glasses more often.

 

[allegansveritatem]

Good work.

But don't forget that "cot" by itself is meaningless in an expression -- it's a function name, not a quantity you can subtract 2 from. So your final answer must include alpha as the argument of each cot.

And, again, there are different valid answers depending on what you call simple. One that hasn't been mentioned is that you could multiply the numerator and denominator by [MATH]\tan\alpha[/MATH] if you like tan more than cot.

Right. I omitted the alpha in the interest of streamlining the process, I guess. But I am aware that it a sin or a tan is exactly nothing without a theta.

 

[allegansveritatem]

Can you please use equal signs. The problem is if you don't then the beginning does not equal the end--or at least you can't conclude that.

Did I really say to convert to sines and cosines? I need to use my eye glasses more often.

OK....but sometimes the expressions don't equal one another, in which case I will make it clear what I mean.

 

[Steven G]

OK....but sometimes the expressions don't equal one another, in which case I will make it clear what I mean.

Fine, but when they do it is customary to use an equal sign.