Understanding a Math Equation Step by Step

[quicksilver123]

Homework Statement

my book shows:

(-1/(1-((b+acosx)/(a+bcosx))^2)^1/2) * ((a+bcosx)(-asinx)-(b+acosx)(-bsinx))/(a+bcosx)^2

=
(a^2+b^2cos^2(x)-b^2-a^2cos^2(x))^(-1/2)
*
((a^2-b^2)sinx/(| a+bcosx |))

I'm having a hard time understanding how they did this.

 

[FactChecker]

Please proof read your equation. I see a variable c on the left side and a variable A on the right that is not matched on the other side. What are they?

 

[quicksilver123]

Edited. Thanks for the heads up, "c" was a typo

 

[Kara386]

Homework Statement

my book shows:

(-1/(1-((b+acosx)/(a+bcosx))^2)^1/2) * ((a+bcosx)(-asinx)-(b+acosx)(-bsinx))/(a+bcosx)^2

=
(a^2+b^2cos^2(x)-b^2-a^2cos^2(x))^(-1/2)
*
((a^2-b^2)sinx/(| a+bcosx |))

I'm having a hard time understanding how they did this.

It's really hard to read your equation. Do you know how to use LaTeX? There's a tutorial here: https://www.physicsforums.com/help/latexhelp/
But in the meantime, is this what you meant?

$\frac{-1}{\sqrt{1-(\frac{b+a\cos(x)}{a+b\cos(x)})^2}} \frac{(a+b\cos(x))(-a\sin(x))-(b+a\cos(x))(-b\sin(x))}{(a+b\cos(x))^2} = \frac{1}{\sqrt{a^2+b^2\cos^2(x)-b^2-a^2\cos^2(x)}} \frac{(a^2-b^2)\sin(x)}{|a+b\cos(x)|}$

Admittedly that's quite tricky to format, brackets don't enclose the fraction and that type of thing, someone might have suggestions on how to improve on it.

 

[quicksilver123]

Thanks!

Actually this is what I meant. I copy and pasted and edited your latex script.
Mod note: Changed from inline tex (## tags)to standalone tex ($$ tags) to make everything larger and easier to read.
$$\frac{-1}{\sqrt{1-(\frac{b+a\cos(x)}{a+b\cos(x)})^2}} \frac{(a+b\cos(x))(-a\sin(x))-(b+a\cos(x))(-b\sin(x))}{(a+b\cos(x))^2} = \frac{1}{\sqrt{a^2+b^2\cos^2(x)-b^2-a^2\cos(x)^2}} \frac{(a^2-b^2)sin(x)}{|a+bcos(x)|} $$Actually we have the same equation despite my edits haha. It just wasn't displaying correctly with my phone being vertical.

Anyway, that's what I need help with.

 

[Mark44]

Thanks!

Actually this is what I meant. I copy and pasted and edited your latex script.

$\frac{-1}{\sqrt{1-(\frac{b+a\cos(x)}{a+b\cos(x)})^2}} \frac{(a+b\cos(x))(-a\sin(x))-(b+a\cos(x))(-b\sin(x))}{(a+b\cos(x))^2} = \frac{1}{\sqrt{a^2+b^2\cos^2(x)-b^2-a^2\cos(x)^2}} \frac{(a^2-b^2)sin(x)}{|a+bcos(x)|}$Actually we have the same equation despite my edits haha. It just wasn't displaying correctly with my phone being vertical.

Anyway, that's what I need help with.

I would start with the left side to see if I end up with what's on the right. For starters, combine the two terms in the radical, and rewrite that whole factor in the form $\frac{\sqrt{\text{something}}}{\sqrt{\text{something else}}}$.
Next expand the products in the numerator of the 2nd factor -- some of the terms will probably drop out.

 

[quicksilver123]

$$\frac{-1}{{1-}\frac{\sqrt{b^{2}+2abcosx+a^{2}cos^{2}x}}{\sqrt{a^{2}+2abcosx+b^{2}cos^{2}x}}}\cdot \frac{-a^{2}sinx+b^{2}sinx}{a^{2}+2abcosx+b^{2}cos^{2}x}$$That's where I am. Obviously I can factor the a^2 and b^2 and the sin on the right numerator but uh... stuck.

Oh, and this handy little tool for latex previews: https://www.codecogs.com/latex/eqneditor.php

 

[Mark44]

$$\frac{-1}{{1-}\frac{\sqrt{b^{2}+2abcosx+a^{2}cos^{2}x}}{\sqrt{a^{2}+2abcosx+b^{2}cos^{2}x}}}\cdot \frac{-a^{2}sinx+b^{2}sinx}{a^{2}+2abcosx+b^{2}cos^{2}x}$$That's where I am. Obviously I can factor the a^2 and b^2 and the sin on the right numerator but uh... stuck.

Oh, and this handy little tool for latex previews: https://www.codecogs.com/latex/eqneditor.php

There'a a preview button here on the lower right.

How did you get this?
$$\frac{-1}{{1-}\frac{\sqrt{b^{2}+2abcosx+a^{2}cos^{2}x}}{\sqrt{a^{2}+2abcosx+b^{2}cos^{2}x}}}$$
You don't show the intermediate steps, but it looks like you did something like this: $\sqrt{1 + x} = \sqrt{1} +\sqrt{x}$, which is incorrect.

In my previous post I said this:

For starters, combine the two terms in the radical, and rewrite that whole factor in the form $\frac{\sqrt{\text{something}}}{\sqrt{\text{something else}}}$.

In case I wasn't clear, I was talking about the radical in the denominator of the first factor on the left side.

 

[quicksilver123]

That's just a typo. The entire fraction is meant to be under a square root.

 

[Mark44]

That's just a typo. The entire fraction is meant to be under a square root.

Your answer here doesn't make sense to me.
Here's part of the left side, from post #5
$$\frac{-1}{\sqrt{1-(\frac{b+a\cos(x)}{a+b\cos(x)})^2}}$$

In post #7 you show this (again only showing part of the left side):
$$\frac{-1}{{1-}\frac{\sqrt{b^{2}+2abcosx+a^{2}cos^{2}x}}{\sqrt{a^{2}+2abcosx+b^{2}cos^{2}x}}}$$

Is this what you really meant to write?
$$\frac{-1}{\sqrt{1 - \frac{\text{stuff}}{\text{other stuff}}}}$$

If so, what I said before was this -- inside that radical combine the 1 and the fraction into a single fraction. Don't keep it as a difference of two terms.

 

[quicksilver123]

Yes that is what I wanted to write. Everything under the square root all at once.

 

[quicksilver123]

Any help?

 

[Mark44]

Any help?

What do you call this?

If so, what I said before was this -- inside that radical combine the 1 and the fraction into a single fraction. Don't keep it as a difference of two terms.

 

[quicksilver123]

See the attached photo for my issue.

 

[Mark44]

View attachment 107674 Any help?

The fourth line up from the bottom looks OK, although there's no good reason to expand the $(a + b\cos(x))^2$ part in the denominator inside the radical.

You have a mistake in the third line up from the bottom. I didn't look at the other factor on the left side.

Also, you have something like this: $$\frac{-1}{\sqrt{\frac{xxx}{yyy}}}$$
This part could be simplified to this:
$$-\sqrt{\frac{yyy}{xxx}}$$

 

[SammyS]

View attachment 107674 Any help?

You don't need to expand so much stuff .

Consider: $\ (a+b\cos(x))^2-(b+a\cos(x))^2 \$

That's a difference of squares thus: $\ (a+b\cos(x)-(b+a\cos(x))\,)(a+b\cos(x)+b+a\cos(x)) \$ .

With some rearranging: $\ ((a-b) - (a-b)\cos(x))((a+b)+(a+b)\cos(x)) \$

$((a-b) (1-\cos(x))((a+b)(1+\cos(x)) \$ and so forth.

 

[Ray Vickson]

You don't need to expand so much stuff .

Consider: $\ (a+b\cos(x))^2-(b+a\cos(x))^2 \$

That's a difference of squares thus: $\ (a+b\cos(x)-(b+a\cos(x))\,)(a+b\cos(x)+b+a\cos(x)) \$ .

With some rearranging: $\ ((a-b) - (a-b)\cos(x))((a+b)+(a+b)\cos(x)) \$

$((a-b) (1-\cos(x))((a+b)(1+\cos(x)) \$ and so forth.

In this case one gets a much simpler final answer by NOT using $A^2-B^2 = (A-B)(A+B)$, but, instead, just expanding out $A^2 =(a+b\,\cos x)^2$ and $B^2 = (b+a\,\cos x)^2$ and cancelling some things.