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Trig question

  1. Oct 17, 2016 #1
    1. The problem statement, all variables and given/known data

    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.
     
    Last edited: Oct 17, 2016
  2. jcsd
  3. Oct 17, 2016 #2

    FactChecker

    User Avatar
    Science Advisor
    Gold Member

    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?
     
  4. Oct 17, 2016 #3
    Edited. Thanks for the heads up, "c" was a typo
     
  5. Oct 18, 2016 #4
    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.
     
  6. Oct 18, 2016 #5
    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.
     
    Last edited by a moderator: Oct 18, 2016
  7. Oct 18, 2016 #6

    Mark44

    Staff: Mentor

    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.
     
  8. Oct 18, 2016 #7
    $$\frac{-1}{{1-}\frac{\sqrt{b^{2}+2abcosx+a^{2}cos^{2}x}}{\sqrt{a^{2}+2abcosx+b^{2}cos^{2}x}}}\cdot \frac{-a^{2}sinx+b^{2}sinx}{a^{2}+2abcosx+b^{2}cos^{2}x}$$


    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
     
  9. Oct 18, 2016 #8

    Mark44

    Staff: Mentor

    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:
    In case I wasn't clear, I was talking about the radical in the denominator of the first factor on the left side.
     
  10. Oct 18, 2016 #9
    That's just a typo. The entire fraction is meant to be under a square root.
     
  11. Oct 18, 2016 #10

    Mark44

    Staff: Mentor

    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.
     
  12. Oct 18, 2016 #11
    Yes that is what I wanted to write. Everything under the square root all at once.
     
  13. Oct 18, 2016 #12
    IMG_2104.JPG Any help?
     
    Last edited: Oct 18, 2016
  14. Oct 18, 2016 #13

    Mark44

    Staff: Mentor

    What do you call this?
     
  15. Oct 18, 2016 #14
    See the attached photo for my issue.
     
  16. Oct 18, 2016 #15

    Mark44

    Staff: Mentor

    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:
    $$-\sqrt{\frac{yyy}{xxx}}$$
     
  17. Oct 19, 2016 #16

    SammyS

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    Staff Emeritus
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    Homework Helper
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    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.
     
  18. Oct 19, 2016 #17

    Ray Vickson

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    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.
     
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