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Difficult Trig Identity (no double ange, power reducing formulas)

  1. Nov 27, 2008 #1
    1. The problem statement, all variables and given/known data

    sin^3(x) - cos^3(x) / sin(x) + cos(x) = csc^2(x) -cot(x) - 2cos^2(x) / 1 - cot^2(x)

    2. Relevant equations



    3. The attempt at a solution

    I have attached[​IMG] one of my many attempts. Any input?
     
  2. jcsd
  3. Nov 27, 2008 #2

    Mark44

    Staff: Mentor

    First off, your identity should be written like so, with added parentheses:
    (sin^3(x) - cos^3(x) ) / (sin(x) + cos(x)) = (csc^2(x) -cot(x) - 2cos^2(x) )/ (1 - cot^2(x))
    The parentheses I added make it clear what's in the numerator and denominator on each side.

    Second, going from the second line of your work to the third there's a mistake.
    You have sin x /(sin x + cos x) in the 2nd line, and 1/(1 + cot x) in the third line.
    Apparently you "cancelled" sin x from the top and bottom, but even so, I still can't figure out how you got the cot x term in the denominator.

    Also in the 2nd line, you have cos x / (sin x + cos x), from which you got 1/(x + 1). There is no way you can justify this.

    These mistakes nullify anything on the 3rd and following lines.

    If I were to prove this identity, I would start on the right-hand side, and convert all of the terms into sine and cosine terms, and try to make it look like the left hand side.
     
  4. Nov 27, 2008 #3
    Thank you for the prompt input. I did make a mistake trying that way. I took your advice and started on the right hand side. Here is where I am at.

    [​IMG]


    [​IMG]
     
  5. Nov 27, 2008 #4

    Mentallic

    User Avatar
    Homework Helper

    On the end of your first page, you cancelled the sin2x from the numerator, but not from the denominator?

    [tex]\frac{(\frac{1-cosxsinx-2cos^2xsin^2x}{sin^2x})}{(\frac{sin^2x-cos^2x}{sin^2x})} \neq \frac{1-cosxsinx-2cos^2xsin^2x}{(\frac{sin^2x-cos^2x}{sin^2x})}[/tex]
     
  6. Nov 28, 2008 #5

    Mark44

    Staff: Mentor

    Alex1067,
    Before doing anything more, could you verify that the problem you posted is exactly as it appears in your book or homework assignment. If you have one sign wrong, it will make for a lot of work for no purpose.
    Thanks...
     
  7. Nov 28, 2008 #6

    Mentallic

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    I've tested a few values of x and it works for all, so it must be accurately transcripted.
     
  8. Nov 28, 2008 #7
    Yes! i managed to prove it.:smile:

    you start off by
    (sin³-cos³)/(sinx+cosx) x (1-cot²x)/(1-cot²x)

    where by:(1-cot²x) is equal to (sin²x-cos²x)/sin²x

    i will be off this thread if you need help just add me in your msn.
    icystrike@physicsforums.com
     
    Last edited: Nov 28, 2008
  9. Nov 28, 2008 #8
    I have still not solved the trig Identity. I gave up on it for the time being and completed two other ones. If anyone has a solution to the first posted identity. Please let me know! Here they are if anyone is interested:


    [​IMG]


    [​IMG]
     
  10. Nov 28, 2008 #9

    gabbagabbahey

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    Gold Member

    I would concentrate on the leftt hand side of the identity. Start by looking at the numerator-- you have a difference of two cubes-- which you can easily factor using [itex]a^3-b^3=(a-b)(a^2+ab+b^2)[/itex]...

    [tex]\Rightarrow \frac{\sin^3x-\cos^3x}{\sin x+\cos x}= \frac{(\sin x -\cos x )(\sin^2x+\sin x \cos x +\cos^2x)}{\sin x+\cos x}= \frac{(\sin x -\cos x )(1+\sin x \cos x)}{\sin x+\cos x}[/tex]

    Now just multiply both the numerator and denominator by [tex]\frac{\sin x -\cos x}{\sin^2x}[/tex] and simplify...
     
  11. Nov 28, 2008 #10
    Can you multiply just one side of the identity by something? I thought it would have to be a form of one?
     
  12. Nov 28, 2008 #11

    gabbagabbahey

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    Gold Member

    That's why you multiply both the numerator and denominator by the same factor...after all; [tex]\frac{a}{b}\cdot \frac{f(x)}{f(x)}=\frac{a}{b}[/tex].

    [tex]\Rightarrow \frac{(\sin x -\cos x )(1+\sin x \cos x)}{\sin x+\cos x}=\frac{(\sin x -\cos x )(1+\sin x \cos x)\cdot \frac{\sin x -\cos x}{\sin^2x}}{(\sin x+\cos x)\cdot \frac{\sin x -\cos x}{\sin^2x}}[/tex]
     
  13. Nov 28, 2008 #12
    Ok thank you. I will try and that and see how it goes!
     
  14. Nov 28, 2008 #13
    [​IMG]

    Here is the step by step solution.
    I hope that you will refer it as reference only.
     
  15. Nov 29, 2008 #14
    Thank you. I did the same thing you did after your original hint of using (1-cot^2(X)).
     
  16. Nov 29, 2008 #15

    Mentallic

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    For your 2nd question. You have 2 pages of swimming through a sea of uncertainty. Go back to the 5th line [tex]LHS=\frac{1+cos^2x}{sin^2x}[/tex] and use the identity that [tex]cos^2x=1-sin^2x[/tex].

    You might find that the question could be solved quite simply and quickly :smile:
     
  17. Nov 29, 2008 #16

    gabbagabbahey

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    Your step by step solution is not only contrary to forum policy; but also contains errors. Please refrain from posting full solutions in the future.

    What happened to the [itex]\sin x +\cos x[/itex] in the denominator? Did you divide it into [itex]\sin^2-\cos^2[/itex]?...If so, you should get [itex](\sin^3 x - \cos^3 x)(\sin x - \cos x)[/itex] for your numerator, not [itex](\sin^3 x - \cos^3 x)(\sin x + \cos x)[/itex]...Your next line seems to have this corrected though; so perhaps it was a typo.
     
  18. Nov 29, 2008 #17
    yup. i will refrain from doing that (:
    tats typo btw.
     
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