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Verifying TrigIdentity

  1. Dec 20, 2009 #1
    1. The problem statement, all variables and given/known data
    Verify the identity:

    csc(x) + sec(x) = cot(x) + tan(x)
    sin(x) + cos(x)


    2. Relevant equations

    sin^2 + cos^2 = 1
    tan^2 + 1 = sec^2
    1 + cot^2 = csc^2



    3. The attempt at a solution

    First i changed everything to sin and cos to try and make it more clear. Let me know if I should bypass this:

    1/sin(x) + 1/cos(x) = cot(x) + tan(x)
    sin(x) + cos(x)

    I'm not really sure where to go from here...should i try to make a common denominator?
     
  2. jcsd
  3. Dec 20, 2009 #2

    rock.freak667

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    Multiply the numerator and denominator by sin(x)cos(x)
     
  4. Dec 20, 2009 #3
    Thanks rock.freak!

    Ok i did that and got:

    cos(x) + sin(x)
    sin(x)^2 + cos(x)^2


    does that look right?
     
  5. Dec 20, 2009 #4
    Or is it this?

    sin(x) + cos(x)
    sin(x)^2cos(x) + sin(x)cos(x)^2
     
  6. Dec 20, 2009 #5

    Mentallic

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    It's this one. The other is obviously not true since [itex]sin^2x+cos^2x=1[/itex] so the denominator would become 1.

    Don't expand out the denominator, instead, see if you can cancel any common factors.
    After this step, it should be easier to work on the right-hand side to try get the left-hand side.
     
  7. Dec 20, 2009 #6
    Thanks Mentallic.

    So I kind of looked at the denominator like this (a^2*b)(a*b^2).

    So I should be able to pull out one "a" and one "b", right?

    Does this seem right?

    sin(x) + cos(x)

    sin(x)cos(x)(sin(x)^2 + cos(x)^2)
     
  8. Dec 20, 2009 #7

    Mentallic

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    Actually the denominator is more like [itex]a^2b+b^2a[/itex].

    Now you can factor out [itex]ab[/itex] to get [itex]ab(a+b)[/itex]. Do you see how this works?

    For the denominator it should be [itex]sinx.cosx\left(sinx+cosx\right)[/itex] so do you see what can be cancelled in that fraction?
     
  9. Dec 20, 2009 #8
    Thanks a lot Mentallic.

    Yes, I see what you did on the factoring. I did that too quickly.

    So with the new denominator, it would give me:

    sin(x) + cos(x)
    sin(x).cos(x) (sin(x) +cos(x))


    It seems like you can cancel out the sin(x) + cos(x) of the numerator with the same thing in parenthesis in the denominator, but am I wrong? Can't you only cancel out things that are multiplied or divided, not attached through addition?

    If so, I'm lost on what to cancel out.
     
  10. Dec 20, 2009 #9
    It looks like there was no need to distribute sinxcosx in the denominator at the beginning; just better to have left it like (sinx + cosx)sinxcosx.

    Yes, sinx + cosx will cancel, then you have 1/(sinx*cosx). Then use the identity 1 = sin2x + cos2x for the numerator.
     
  11. Dec 20, 2009 #10
    Thanks Bohrok.

    I'm still a little confused about how that would verify the identity

    Are you saying to set up the fraction like this now?

    sin^2x + cos^2x
    sinx*cosx

    I see the exponent and denominator would cancel out, but how would that give me cot(x) + tan(x) as the final solutions
     
  12. Dec 20, 2009 #11

    Mentallic

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    Yes you're allowed to cancel the [itex]sinx+cosx[/itex] from the numerator and denominator. They're both the same factor.

    e.g. if we have a fraction such as:

    [tex]\frac{xy}{x}[/tex]
    we can cancel x from both the numerator and denominator because they're common factors.
    This rules works for whatever x may stand for, such as something more complicated like [itex]sina+cosb[/itex]
    Then the fraction would be: [tex]\frac{(sina+cosb)y}{sina+cosb}[/tex] and now you can cancel in the same fashion.


    Ok so after you cancel the [itex]sinx+cosx[/itex] you will have [tex]\frac{1}{sinx.cosx}[/tex].
    It is a little hard to see what you could do with this to try get the RHS, so it would be easier to work on the RHS from this point.
     
  13. Dec 20, 2009 #12
    Then simplify the fraction using the fact that (a + b)/c = a/c + b/c. You could work on the right hand side as Mentallic said so you can see how it should work out.
     
  14. Dec 20, 2009 #13
    Thank you guys for being so kind and giving me your time!! I think I can get the rest from here.
     
  15. Dec 20, 2009 #14

    Mentallic

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    No problem, good luck!
     
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