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Trig Identities

  1. Jan 17, 2005 #1
    Hi... I need help solving this problem. I don't know what to do...

    Proove that

    [tex]\frac{cos^2\theta - sin^2\theta}{cos^2\theta+sin\theta cos\theta} = 1 - tan\theta[/tex]

    I tried to cross out cos^2 on the top with the one at the bottom... also tried messing around with the values (cos^2x = 1 - sin^2x) etc... but nothing is working. I'm kind of lost and would appreciate any help... Thanks
     
  2. jcsd
  3. Jan 17, 2005 #2
    First if you are given [tex]\frac{cos^2\theta - sin^2\theta}{cos^2\theta+sin\theta cos\theta} = 1 - tan\theta[/tex] factor the numerator (its a difference of two perfect squares). Then factor the denominator (can you see a common term?) Cancel like terms. Divide through and you will get your result.

    Hint: [tex] a^2 - b^2 = (a-b)(a+b) [/tex]
     
    Last edited: Jan 17, 2005
  4. Jan 17, 2005 #3
    Cross out with [tex]cos^2(x)[/tex] also works.


    [tex]\frac{cos^2(x)-sin^2(x)}{cos^2(x)}[/tex]
    [tex]\frac{cos^2(x)+sin(x)cox(x)}{cos^2(x)}[/tex]

    When you solve these, the answer comes out very nicely
     
    Last edited: Jan 17, 2005
  5. Jan 17, 2005 #4
    maybe it's because it's only been a little over a week from the xmas break, but even after factoring, I don't know the next step... how can i cancel out stuff still in the brackets? And for crossing it out... I don't really get what happened. What did you cross out exactly, and why is the fraction flipped?

    factored...
    [tex]\frac{(cos\theta + sin\theta) (cos\theta - sin\theta)}{(cos\theta+sin\theta)(cos\theta)}[/tex]
     
    Last edited: Jan 17, 2005
  6. Jan 17, 2005 #5
    In the denominator, inside the brackets you get [tex](cos\theta+sin\theta)[/tex] because before factoring you had [tex]cos^2\theta[/tex]. Now do some cancelling out and you will get the answer.
     
  7. Jan 17, 2005 #6
    yah, i just caught my mistake ... but im still not getting 1-tan theta...
    I end up with cos theta - sin theta over cos theta after cancelling...
     
  8. Jan 17, 2005 #7
    ok so you see [tex]\frac{(cos\theta + sin\theta) (cos\theta - sin\theta)}{(cos\theta+sin\theta)(cos\theta)}[/tex] [tex](cos\theta+sin\theta)[/tex] both in the numerator and denominator. So cancel that out and you are left with:

    [tex]\frac{(cos\theta - sin\theta)}{(cos\theta)}[/tex]

    Simplify this and you get the answer
     
    Last edited: Jan 17, 2005
  9. Jan 17, 2005 #8
    [tex]\frac{cos\theta-sin\theta}{cos\theta}=\frac{cos\theta}{cos\theta}-\frac{sin\theta}{cos\theta}[/tex]
     
  10. Jan 17, 2005 #9

    Curious3141

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    What is the definition of [itex]\tan \theta[/itex] in terms of [itex]\sin \theta[/itex] and [itex]\cos \theta[/itex] ?
     
  11. Jan 17, 2005 #10
    mmmannnn.... I think I need another vacation. Thanks for the help =)
     
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