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A Few Simple Calculus Problems

  1. Mar 1, 2004 #1
    Good morning folks, I have been recently stumped by a few of my son's homework calculus problems and was hoping for a little bit of help. I'm sure they are simple, the thing is it has been 20 years since I took this.

    2cosx=sinx
    sin(pi)/2x=1
    sin^2x+sin=1 (find zeros)
    coscot=2cos
    tan^2/cos(90-x)
    (sinx^x-cosx^4)/(sinx^2-cosx^2)=1
    cosx(secx-cosx)

    I appreciate all your help :smile:
     
  2. jcsd
  3. Mar 1, 2004 #2
    1. 2cosx = sinx
    Divide through by cosx
    2 = tanx
    x = arctan2

    2. sin(pi)/(2x) = 1
    2x = sin(pi)
    sin(pi) is equal to 0, so
    x = 0

    3. (sinx)^2 + sinx - 1 = 0
    Factor using the quadratic formula
    [tex]\sin x = \frac{-1 \pm \sqrt{1 - 4*1*(-1)}}{2}[/tex]
    Then just take the arcsin.

    4. I don't understand this notation.

    5. We're short an equals sign on a right-hand sign here. But this identity might help:
    cos[Pi/2 - x] = sin[x]

    6. & 7. You should probably clean up the notation in these ones before we attempt them. It's a little ambiguous and different interpretations really change the problems. And there's no equals sign in 7.

    cookiemonster
     
  4. Mar 9, 2004 #3
    For question 4:

    "coscot=2cos"

    I assume that would be (cosx)(cotx) = 2cosx"?

    In which case

    cot x = 2
    (1/tan x) = 2
    1 = 2 tanx
    tan x = 1/2
    x = arctan 1/2

    Helpful?

    Also, if 7 was supposed to be "cosx = (secx-cosx)"

    Then:

    cos x = sec x - cosx
    sec x = 0
    (cos x)^-1 = 0

    Undefined ( divide by zero)

    But I'm guessing thats not what you meant.

    Probably kinda late now anyway, but oh well.
     
  5. Mar 9, 2004 #4
    For number 6. I am assuming you mean sinx^4-cosx^4? If so, then (sinx^2+cosx^2)(sinx^2-cosx^2)/(sinx^2-cosx^2)
    sinx^2+cosx^2
    1=1
     
  6. Mar 10, 2004 #5
    Just remember that the tangent has a period of [tex]\pi[/tex], so [tex]x = \arctan{2} + n\pi[/tex]. It's also good to check that [tex]\cos{x} \neq 0[/tex] before dividing with it...
     
    Last edited: Mar 10, 2004
  7. Sep 22, 2004 #6
    Need some help..

    Can someone please help me with this one?

    cos2x = 2 cos x sin x :uhh:
     
  8. Sep 22, 2004 #7
    It's equivalent to cos(2x) = sin(2x), or 1 = sin(2x)/cos(2x)...
     
  9. Sep 22, 2004 #8
    help..

    I need to find 4 angles..
     
  10. Sep 22, 2004 #9

    Zurtex

    User Avatar
    Science Advisor
    Homework Helper

    Generally best to make your own thread.

    [tex]\sin 2x \equiv 2 \cos x \sin x[/tex]

    So as stated above your problem is the same as:

    [tex]\sin 2x = \cos 2x[/tex]

    Which is the same as:

    [tex]\frac{\sin 2x}{\cos 2x} = 1[/tex]

    Now you should remember a simple identity about sin over cos which makes this really simple.
     
  11. Sep 22, 2004 #10
    I`ll that.. Thanks :smile:
     
  12. Sep 22, 2004 #11
    Thanks

    I got it right.. thanks a lot :rofl:
     
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