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Trig identities

  1. Feb 26, 2012 #1
    1. The problem statement, all variables and given/known data

    (3/5)cos2x + (3/5)sin2x



    3. The attempt at a solution

    I would think the answer would be 6/5, but it looks like the book is saying 3/5. I had a similar problem to this the other day and I tried finding it in my history but I couldn't.
     
  2. jcsd
  3. Feb 26, 2012 #2
    I think you just made a careless mistake in factorizing...
    Try again.
     
  4. Feb 26, 2012 #3
    Well, are you saying the book is wrong? That it's not 3/5?
     
  5. Feb 26, 2012 #4
    I mean you are wrong...
    show your working on eliminating the sin and cos
     
  6. Feb 26, 2012 #5
    ok, I take it

    x cos^2 + x sin^2 = x, not 2x, that's what i needed to know.
     
  7. Feb 26, 2012 #6
    Suppose you have:

    [tex] x \cdot a + x \cdot b = c [/tex]

    Then you clearly can't group the a and b together. However, this is what you can do:

    [tex]
    x\cdot(a+b) = c
    [/tex]

    In your case:

    [tex]
    \frac{3}{5} \cdot (cos^{2}(x) + sin^{2}(x)) =
    [/tex]
    [tex]
    \frac{3}{5} \cdot 1
    [/tex]
     
  8. Feb 26, 2012 #7

    Mark44

    Staff: Mentor

    Review your trig identities!
     
  9. Feb 26, 2012 #8
    when i saw that mark commented on this post, i knew he would yell at me for asking an elementary question. mark, you need to find something better to do with your time other than yell at people for trying to learn. i spend as much time reviewing trig identities as i see necessary.
     
  10. Feb 26, 2012 #9

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Rather than Mark suggesting that you study trig identities, (Really, I don't see at all how that was yelling at you.) I'm going to point out that the statement above is nonsense .

    You need to have an argument to go with a function.

    The following is true:

    x cos2(θ) + x sin2(θ) = x​

    How can you show it's true?

    x cos2(θ) + x sin2(θ) = x (cos2(θ) + sin2(θ)) = x (1) = x .

    Similarly, for the problem you posed to begin this thread, simply factor out (3/5) & use one of the Pythagorean trig. identities.

    When you write sin2x , that's not x times the square of the sine, whatever that might mean.
     
  11. Feb 26, 2012 #10

    Mark44

    Staff: Mentor

    "as I see necessary." - In every post of yours that I have seen, you have gotten stumped on some very elementary algebra property or trig identity in some explanation you are reading in a math book. I have yet to see you asking a question about a problem you are actually working on - your problems seem to be in understanding what to many would be a clear explanation.

    As I said before, I think that what you are doing is to be commended, but it seems to me that until you have a firm grip on the algebra and trig fundamentals that form the foundation of calculus, you are wasting your time. I would never "yell" at people for trying to learn, but if they are obviously not prepared to study a particular area of mathematics, I would point that out, and advise them to strengthen the areas they are weak in, and that's what I've been doing with you.
     
  12. Feb 26, 2012 #11

    Mark44

    Staff: Mentor

    = (3/5)(cos2(x) + sin2(x)) = (3/5)(1) = 3/5
    This problem is nothing more than an application of the distributive property and the identity cos2(x) + sin2(x) = 1.

    If you have any hope of understanding calculus, you NEED to get squared away on the basic stuff.
     
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