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Vector Problem

  1. Sep 24, 2007 #1
    1. The problem statement, all variables and given/known data

    Sin(squared) theta + Cos(squared) theta = 1

    I'm new to vectors and am having difficulty solving this, any help would be greatly appreciated. Thanks.
  2. jcsd
  3. Sep 24, 2007 #2


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    There are multiple ways to do this... what are you supposed to start with... is this regarding dot product by any chance?
  4. Sep 24, 2007 #3
    Remember you can write a unitary vector as:

    cos(theta) i + sin(theta) j

    Shouldn't be too hard to go from there.
  5. Sep 24, 2007 #4


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    This has nothing to do with vectors! It's the Pythagorean theorem!
  6. Sep 24, 2007 #5
    My instructor said solve it any way possible. Could someone show me how to work it out using the pythagorean theorem, I'm still very lost at this.
  7. Sep 24, 2007 #6


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    Suppose you have a right triangle with hypoteneuse 1, and an angle theta. What are the lengths of the other two sides?
  8. Sep 24, 2007 #7
    we just learned angle theta today, and as ignorant as this sounds what exactly is an angle theta. I understand where it is on the right triangle, but does it have an exact amount of degrees and how does it relate to the problem in helping to solve it.
  9. Sep 24, 2007 #8
    okay so
    Sin theta = y/r and
    Cos theta = x/r

  10. Sep 24, 2007 #9


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    yeah... so y = rsintheta. x = rcostheta

    we know that by the pythagorean theorem x^2 + y^2 = r^2. sub in the values of x and y in terms of theta...
  11. Sep 24, 2007 #10
    its been a while since I used the pythagorean theorem so...

    x^2+y^2 = 1

    now where would I go from here since there are two variable to solve for? lol I'm just forgetting all the basics.
  12. Sep 24, 2007 #11
    If I remember correctly, angles play a role in solving for two variables, right
  13. Sep 24, 2007 #12
    ok I think I got it,

    Sin^2 theta = (.71/1)^2
    Cos^2 theta = (.71/1) ^2

    (.71/1)^2 + (.71/1)^2 = 1.0082 (about 1)

    thanks for the help:)
  14. Sep 24, 2007 #13


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    yeah, but you don't need to assume a particular theta...

    x^2 + y^2 = r^2

    (rcostheta)^2 + (rsintheta)^2 = r^2

    then divide both sides by r^2

    and you get

    (costheta)^2 + (sintheta)^2 = 1
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