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Calculus Summer Assignments

  1. Jul 7, 2005 #1
    I'm taking Calculus next year and over the summer I have some assignments.
    This one is due in a couple of hours, so any help would be appreciated!

    If cos2t=1/3 and *0<_ 2t <_ pie, find cost. t=theta *less than or equal to

    I dont know how to use the identities to help me.
    :confused:
    Please help!
     
  2. jcsd
  3. Jul 7, 2005 #2
    Well....

    [tex]\cos{(2\alpha)} = 2\cos^2{(\alpha)}-1[/tex]

    Plug that in and post when you make progress.
     
  4. Jul 7, 2005 #3
    ohh, costheta=sqrt(6)/3
    ?
     
  5. Jul 7, 2005 #4
    [tex]2\cos^2{(\alpha)}-1 = \frac{1}{3}[/tex]

    [tex]2\cos^2{(\alpha)}=\frac{4}{3}[/tex]

    [tex]\cos^2{(\alpha)}=\frac{2}{3}[/tex]

    Can you finish from here?
     
  6. Jul 7, 2005 #5
    oh yes thanks!

    how about this one.

    I'm not sure how to simplify it down, and how to distribute the ^2 once it has been plugged in.
    x^2 + y^2 +3x=0 when x=rcostheta and y=rsintheta
     
  7. Jul 7, 2005 #6
    Remember that [itex]\sin^2 x + \cos^2 x = 1[/itex]. These questions don't seem to have anything to do with calculus, they just seem to be trigonometry.
     
  8. Jul 7, 2005 #7
    [tex](r\cos{\theta})^2 + (r\sin{\theta})^2+3(r\cos{\theta})=0[/tex]

    [tex]r^2\cos^2{(\theta)}+r^2\sin^2{(\theta)}+3(r\cos{\theta})=0[/tex]

    Do you see the trig identity coming in?
     
    Last edited: Jul 7, 2005
  9. Jul 7, 2005 #8
    the Pythag. Identity? Would you have to plug in rcostheta with the 3x?
     
  10. Jul 7, 2005 #9
    I should have plugged that in earlier. But no, that's not where the identity comes in.

    I'll give you my last hint to this problem.

    [tex]r^2\cos^2{(\theta)}+r^2\sin^2{(\theta)}+3(r\cos{\theta})=0[/tex]

    [tex]r^2(\cos^2{\theta}+\sin^2{\theta})...[/tex]
     
  11. Jul 7, 2005 #10
    OH! thanks!!!!!
     
  12. Jul 7, 2005 #11
    when you distribute the 3, would it be 3rcos3theta? or do you just not distribute the 3 to the cos?
     
  13. Jul 7, 2005 #12
    [tex]3r\cos{(\theta)}[/tex]
     
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