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Another Trig Identity

  1. Oct 9, 2005 #1
    [tex]1 - \frac{\cos^2 \theta}{1 + \sin^2 \theta} = \sin \theta[/tex]

    I can only get to [tex] \frac{2\sin^2 \theta}{1 + \sin^2 \theta}[/tex] and I don't know if that's correct.
     
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  3. Oct 9, 2005 #2

    TD

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    Perhaps you should check the problem again because it doesn't seem correct to me. Try [itex]\theta = 3\pi /2[/itex], the LHS will be 1 while the RHS is -1.
     
  4. Oct 9, 2005 #3

    Fermat

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    It's not an identity!

    Let θ = -π/2

    then lhs = 1 - cos²(-π/2)/(1 + sin²(-π/2))
    lhs = 1 - 0/2
    lhs = 1
    =====

    but rhs = sin(-π/2) = -sin(π/2) = -1
    rhs = - 1
    =======

    Since lhs ≠ rhs, it can't be an identity.
     
  5. Oct 9, 2005 #4
    OMG, I've spent so much time trying to figure it out... I never thought of checking to make sure it was indeed an identity. I guess my teacher made a mistake. Thanks guys.

    (Don't worry, I'll be back with another too soon :tongue2:)
     
  6. Oct 9, 2005 #5

    Fermat

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    I did a quick check to see if it was an identity. After all , you were having trouble !
    It checked out ok for θ = 0 and θ = pi/2, so I did some work on it.
    It was only later on that I plugged the expressions into graphmatica and found out they were actually two different functions !!
     
  7. Oct 10, 2005 #6
    [tex]\frac{1 - \sin \theta}{\cos \theta} + \frac{\cos \theta}{1 - \sin \theta} = 2 \sec \theta[/tex]

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

    I can't seem to get anywhere with these...
     
  8. Oct 10, 2005 #7

    Fermat

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    For the 1st one, cross-multiply the two terms in the lhs and simplify.
     
  9. Oct 10, 2005 #8

    Fermat

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    For the 2nd one, multiply the lhs by (1 - sinθ) and simplify.
     
  10. Oct 10, 2005 #9
    I can only get to [itex]2 \sin \theta + 2 \sin^2 \theta[/itex] for the first after cross multiplying.

    For the second, how can I just multiply by [itex](1 - \sin \theta)[/itex]? I thought you can only multiply by 1?
     
  11. Oct 10, 2005 #10

    arildno

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

    [tex]\frac{1 - \sin \theta}{1 + \sin \theta} =\frac{1-\sin\theta}{1-\sin\theta}\frac{1-\sin\theta}{1+\sin\theta}=\frac{(1-\sin\theta)^{2}}{\cos^{2}\theta}=(sec\theta-tan\theta)^{2}[/tex]
     
  12. Oct 10, 2005 #11
    I see, thanks.
     
  13. Oct 10, 2005 #12

    Fermat

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    Ah, sorry, that should have been, multiply above and below by (1 - sinθ)
     
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