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Easy algebra - Calculus question help - find roots of

  1. Jul 21, 2009 #1
    1: Which of the following is incorrect

    a cos2x = 2-(sin^2) x
    b (sin^2)x + (cos^2)x = 1
    c cos x = sin ( pi/2 -x)
    d sin2x = 2sinxcosx

    I know b is correct, fairly sure c is correct because i know cos is out of phase by pi/2
    have no bloody idea how to determine which of these is wrong

    help please, need to know the rules/trig identities


    2:
    if a = 3/4, then 1/ a+1 = 1/1.75
    HOWEVER i need to know how to do this, without a calculator, we are allowed NO calculator during the exam, and the awnser is something like 0.57632

    3: 1/x - 1/(x-3) = a / x(x-3)
    solve for a,

    i tried multiplying the first side by x-3, so got (x-3)/(x(x-3)) - (x-3)/(x-3)
    (x-3)/(x^2-3x) - 1 = a / (x^2 -3x)
    moving (x^2-3x),
    (x-3) -1 = a
    x -4 =a?????????????

    help pl0xzafd
     
  2. jcsd
  3. Jul 21, 2009 #2
    The easiest way to find the incorrect equations is to pick values at which cosine and sine are easy to evaluate. For example, let x=pi or x=0, and see which equations fail to be true.
    If a=3/4, then
    [tex]
    \frac{1}{a+1} = \frac{1}{3/4+1} = \frac{1}{3/4 + 4/4} = \frac{1}{7/4} = \frac{4}{7}
    [/tex]
    Remember that dividing by a fraction is the same as multiplying by its reciprocal. Can you divide 7 into 4 to get the first few digits of 0.57632?

    Let's hold off on #3 until we get these.
     
  4. Jul 21, 2009 #3

    Sweet, i get number two, i didn't think of turning 1 into 4/4 XD

    as for the first question, I am not allowed to use a calculator at all during the test, so can't really substitute in complex cos/sine stuff (i know simple ones like cos 90= 0 ect..ect..) but nothing to hardcore
     
  5. Jul 21, 2009 #4

    Pengwuino

    User Avatar
    Gold Member

    While you definitely should verify each trigonometric equation, what happens with the first one is at x=0? A nice tip for stuff like this is thinking of the magnitude of sine and cosine. Sin^2, Sin^3, Sine to any power along with Cosine to any power or with any argument (like 2x, 3x, 4x, 10414x) have a maximum magnitude of 1. If you multiply them together (such as Sin(3x)Cos(2x) ), however, this argument doesn't hold.
     
  6. Jul 21, 2009 #5
    Turning 1 into 4/4 is the same process as finding a common denominator. Also, I meant try x=pi/2, not pi. Sorry. And pi/2 is 90 degrees in radians. So what happens when you substitute in x=0 or x=90? If an equation gives something that isn't true, like 0=1 or something else, then the equation is incorrect. If it gives something correct, then it is likely to be a correct equation (but not always though!), so you need to just have the trigonometric equations memorized before a test.
     
  7. Jul 21, 2009 #6
    yeah i just looked up the trigonometric identities on google, and i'm memorizing them all
    i know most of them, but some i missed out! must've been talking during class at highschool lol.

    turns out, well at least i think, question 1, a is wrong
     
  8. Jul 21, 2009 #7
    How do i find the roots of x in the equation

    2x^2 -5x +1

    -

    my attempt, isn't it -5?
     
  9. Jul 21, 2009 #8
    Right, but make sure you know why. If you let x=0, then cos(0)=1, but 2-sin2(0)=2-0=2, which isn't 1. So equation (a) is incorrect.

    For #3, there are a few ways to do this, but a good technique to know is cross-multiplying to find the common denominator between two fractions, i.e.
    [tex]
    \frac{a}{b} \pm \frac{c}{d} = \frac{ad\pm bc}{bd}
    [/tex]
    a,b,c,d can be either numbers, variables like x, or equations such as x-3 in our case.

    So using cross-multiplication,
    [tex]
    \begin{align*}
    \frac{1}{x}-\frac{1}{x-3} &= \frac{a}{x(x-3)} \\
    \frac{(x-3)-x}{x(x-3)} &= \frac{a}{x(x-3)}
    \end{align*}
    [/tex]
    Can you solve for a from here?
     
  10. Jul 21, 2009 #9
    This you should be able to know for sure. Did you substitute -5 back in to see if it works?
    2x2-5x+1=0
    (2x+a)(x+b)=0
    Can you find numbers a and b (which could be negative) so that when you multiply everything out, you get back your quadratic equation? In other words, can you factor the quadratic easily? If not, you will have to use the quadratic equation. If you don't remember it, look it up in your book.
     
  11. Jul 21, 2009 #10

    thanks for that first one,

    -x/x = a/ x(x-3)
    -1 * (x*x - 3x)
    -1(x^2 -3x)
    -x^2 +3x = a
     
  12. Jul 21, 2009 #11
    No! You can't cancel out something out like that. In other words,
    [tex]
    \frac{(x-3)-x)}{x(x-3} \not= \frac{-x}{x}
    [/tex]
    What is the numerator in that equation if you add everything together?
     
  13. Jul 21, 2009 #12
    I think that's my problem, i'm at Uni now and I've just relied on the quadratic equation to get me through those kinds of problems,

    but now, as i've said above we aren't allowed to use calculators so yah.
    reading about quadratics just now XD
     
  14. Jul 21, 2009 #13
    (x-3)-x / (x^2-3x) = a/ (x^2-3x)
    (x-3)-x = a?
     
  15. Jul 21, 2009 #14
    But you don't get -5 when using the quadratic equation, and you should have found two roots, not just one (unless they were imaginary). You need to be able to factor out quadratics in your head or on paper, and you also need to realize when you can't quickly factor it out to just resort to the quadratic equation. Learning to factor can just save some time, but the quadratic equation will always get you the roots.

    Were you able to finish #3?
     
  16. Jul 21, 2009 #15
    No reason to multiply out the denominators though. You could have just multiplied by x(x-3) on both sides to cancel the denominators.

    What does (x-3)-x equal?
     
  17. Jul 21, 2009 #16
    You shouldn't have to memorize all sorts of trig identities; you should be able to manipulate all the trig functions you'll come across in your class using a few basic identities. I can get pretty much anywhere using the sum formula, and sin^2x + cos^2x = 1 (and really fundamental stuff like sinx/cosx = tanx, obviously).

    I remember doing trig identities in my Precalculus class. It was like a drill sergeant giving soldiers draconian amounts of push ups. We manipulated probably hundreds of practice trig identities to get them into the form the question wanted -- and I basically never used anything but those few formulas above to get there.

    If you have trouble getting a trig function into the right form, just evaluate the trig function at pi/6, pi/4, pi/2, pi, or 2pi. They're all easy numbers to evaluate the trig functions at. Good luck on your tests!
     
  18. Jul 21, 2009 #17
    -3 :)

    -

    so using the quadratic formula will give you a different awnser compared to factorization???
     
  19. Jul 21, 2009 #18
    so for example, x^2 +4x -12

    factorizing it, you'd get 6 and -2?
    i didn't use the quadratic formula for that
     
  20. Jul 21, 2009 #19
    Yes, a=-3 is correct. No. The quadratic equation will always give you the two roots, but factorization can get you the same two roots quicker if the quadratic can be factored easily. For example,
    x2-5x+6=0
    (x-3)(x-2)=0
    Now the roots are x=3,2. If you used the quadratic equation, it would take longer and could possibly lead to a mistake.
     
  21. Jul 21, 2009 #20
    right, cheers buddy for all your help
     
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