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Trig Identities

  1. Dec 23, 2014 #1
    Prove the following identity: [tex]\frac{1-tan^2(\theta)}{1-cot^2(\theta)}=1-sec^2(\theta)[/tex]

    By tan/sec identity,

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

    separated the variables

    [tex]\frac {2}{2-csc^2}- \frac {sec^2}{2-csc^2}[/tex]

    There's absolutely nothing useful I can do with either of those terms, so that approach is useless. I also tried putting everything into terms of sin and cos and simplifying to get

    [tex]\frac {sin^2- \frac {2sin^2}{cos^2}}{2sin^2-1}[/tex]

    but that wasn't helpful either. I'm assuming that I need to simplify to -tan^2, but I can't find any way to do that.
     
  2. jcsd
  3. Dec 23, 2014 #2

    ehild

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    Are you sure you copied the problem correctly? It is not an identity.
     
  4. Dec 23, 2014 #3

    Dick

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    It looks like one to me.
     
  5. Dec 23, 2014 #4
    Keep in mind you don't have to start the proof on the left side every time. I always tackle a problem from a simpler looking side myself. So try starting from the right.

    Edit: Got my directions mixed up. Naps are hard.
     
    Last edited: Dec 23, 2014
  6. Dec 23, 2014 #5

    SteamKing

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    Sure, it is.
     
  7. Dec 23, 2014 #6

    SteamKing

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    Hint: use the definition of tangent and cotangent.
     
  8. Dec 23, 2014 #7
    I did start from the left. It looks like I made an algebra mistake, the sine and cosine approach should have left me with

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

    which simplifies to

    [tex]\frac{\left( \tan ^2\right) \left( \cos ^2- \sin ^2\right)}{ \sin ^2- \cos ^2}[/tex]

    still not sure if this is the best approach....
     
  9. Dec 23, 2014 #8
    got my left and rights mixed up :P
     
  10. Dec 23, 2014 #9
    I just did it. Start on the left side, simplify everything to sin and cos. Factor out a negative somewhere in there, then cross multiply, and it all works out.
     
  11. Dec 23, 2014 #10

    ehild

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    Compare the domains. It is an identity except a lot of angles. :)
     
  12. Dec 23, 2014 #11
    [tex]
    \frac{1-tan^2(\theta)}{1-cot^2(\theta)}=1-sec^2(\theta)
    [/tex]

    [tex]\frac{1- \frac{sin^2}{cos^2}}{1- \frac{cos^2}{sin^2}}[/tex]

    [tex]\frac{ \frac {cos^2-sin^2}{cos^2}}{\frac {sin^2-cos^2}{sin^2}}[/tex]

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

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

    [tex]-tan^2[/tex]

    [tex]1-sec^2[/tex]

    thanks, I missed the factor out a negative part. Is there a better way of solving this? It seems a bit esoteric....
     
  13. Dec 23, 2014 #12
    There are literally an infinite number of ways to solve any trig identity, but some are "better", more logical, or quicker than others. The beautiful solutions are the ones that are esoteric. That being said, factoring out the negative is what I saw on my first try.
     
  14. Dec 23, 2014 #13
    I'm teaching myself trig so that I can take calc 1. If this is the method of simplifying a trig problem that would be used in calculus, then I am satisfied; otherwise, I need to practice other solutions. Is it common in calculus to break things into sine/cosine or are there better approaches?
     
  15. Dec 23, 2014 #14

    Fredrik

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    ..or just ##-\tan^2\theta##. So now it's sufficient to show that ##1-\sec^2\theta=-\tan^2\theta##.

    That's how I do these problems. But you might be able to save some time if you memorize some identities that involve those other functions.
     
    Last edited: Dec 23, 2014
  16. Dec 23, 2014 #15

    SteamKing

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    It depends. In this case, using the definition of tan and cot led to proving the identity. In calculus, you'll use sin2+cos2 = 1 and several other identities more often than not, which is why this problem is good practice, not only to improve your trig skills, but your algebra skills as well.
     
  17. Dec 23, 2014 #16

    ehild

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    I still claim that it was not identity.
    Two expression containing a variable are identical if they evaluate the same for any value of the variable in the domain stated. There was no restriction for the domain so the two sides must be defined and identical for any value of θ to be identical.

    What happens if θ is pi/2 or pi or pi/4? Are the expressions on both sides defined?

    In tests, it is a frequent trick question: is (x2-1)/(x-1) = x+1 an identity? The right answer is NOT, as the left side is not defined at x=1, while the right side is.
     
  18. Dec 24, 2014 #17

    SteamKing

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    There are a couple of well known identities which don't meet your criteria, yet they are still called identities. To wit:

    1 + tan2 θ = sec2 θ and
    1 + cot2 θ = csc2 θ

    Although these expressions and others are called identities, and they are not defined for certain values of θ, they nevertheless hold true for all other values of θ.

    http://en.wikipedia.org/wiki/List_of_trigonometric_identities
     
  19. Dec 24, 2014 #18

    ehild

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    They are all right as the domains of both sides are the same. It is not true for the original problem.
     
  20. Dec 27, 2014 #19

    lurflurf

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    ^Most books say an equation is an identity if it is true for all values for which both sides are true. Perhaps there are some very bad books that use your definition. Making a big deal about the domain is silly since most books abuse the domain often. With your definition
    $$\sec^2(\theta)=1+\tan^2(\theta)\\
    \text{is an identity while}\\
    \sec^2(\theta)-\tan^2(\theta)=1\\
    \text{is not}$$
    and
    $$1-\dfrac{\sin(\theta)}{\sin(\theta)}=\dfrac{\sin(\theta)}{\sin(\theta)}-1\\
    \text{is an identity while}\\
    \dfrac{\sin(\theta)}{\sin(\theta)}-1=0\\
    \text{is not}$$
    in fact with your definition we often have
    left hand side of identity-right hand side of identity=0
    is not an identity

    for the given problem I would have done

    $$1-sec^2(\theta)=-tan^2(\theta)=-tan^2(\theta)\dfrac{1-\cot^2(\theta)}{1-\cot^2(\theta)}$$
     
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