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Homework Help: Trig Identities

  1. Oct 13, 2005 #1
    I can't get anywhere with these three identities. Any tips?

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

    [tex]\frac{\sin^3 \theta + \cos^3 \theta}{\sin \theta + \cos \theta} = 1 - \sin \theta\cos \theta[/tex]

    [tex](2a\sin \theta\cos \theta)^2 + a^2(\cos^2 \theta - \sin^2 \theta)^2 = a^2[/tex]
     
  2. jcsd
  3. Oct 13, 2005 #2

    Pyrrhus

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    The third one is pretty obvious a (remember the double angle identities!)
    [tex] \sin^{2} \theta + \cos^{2} \theta = 1 [/tex]
    For the second one remember
    [tex] a^3 + b^3 = (a+b)(a^2 -ab + b^2) [/tex]
     
    Last edited: Oct 13, 2005
  4. Oct 13, 2005 #3

    Pyrrhus

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    I was trying to think a simpler way for the first one, but it all occurs to me now is to

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

    then work it out with sines and cosines.
     
  5. Oct 13, 2005 #4
    Thanks for the tips so far. I got second one.

    For the first, I had simplified it to that already and tried sines and cosines but I'll try again.

    For the last, is there any way to get to the answer from

    [tex]2a^2 + a^2 \sin^4 \theta + a^2 \cos^4 \theta[/tex]

    doesn't seem so...
     
    Last edited: Oct 13, 2005
  6. Oct 13, 2005 #5

    Pyrrhus

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    For the last one is basicly applying
    [tex] \sin 2 \theta = 2 \sin \theta \cos \theta [/tex]
    [tex] \cos 2 \theta = \cos^{2} \theta - \sin^{2} \theta [/tex]
    [tex] \sin^{2} 2 \theta + \cos^{2} 2 \theta = 1 [/tex]
     
  7. Oct 13, 2005 #6
    I've never used those identities before, no wonder I didn't know what was going on :uhh:

    Thanks for your help
     
  8. Oct 13, 2005 #7

    Pyrrhus

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    Well, here are their proof
    [tex] \sin (a+b) = \sin a \cos b + \cos a \sin b [/tex]
    [tex] \sin (\theta + \theta) = \sin \theta \cos \theta + \cos \theta \sin \theta [/tex]
    [tex] \sin (2\theta) = 2 \sin \theta \cos \theta [/tex]
    [tex] \cos (a+b) = \cos a \cos b - \sin a \sin b [/tex]
    [tex] \cos (\theta + \theta) = \cos\theta \cos \theta - \sin \theta \sin \theta [/tex]
    [tex] \cos (2\theta) = \cos^{2}\theta - \sin^{2} \theta[/tex]
     
  9. Oct 13, 2005 #8
    Aha! Thanks again :smile:
     
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