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Another trig expression

  1. Aug 26, 2013 #1
    1. The problem statement, all variables and given/known data

    Simplify:

    (sin t + cos t)2
    /
    sin t cos t

    2. Relevant equations

    tan t = sin t/cos t
    cot t = cos t/sin t
    cot t = 1/tan t

    3. The attempt at a solution

    (sin t + cos t)2
    /
    sin t cos t


    sin2 t + 2sin t cos t + cos2 t
    /
    sin t cos t


    (sin t/cos t) + 2 + (cos t/sin t)

    tan t + cot t + 2

    tan t + (1/tan t) + 2


    (tan2 t + 2tan t + 1)
    /
    tan t


    (tan t + 1)2
    /
    tan t

    The above answer look wrong. I don't know what else to do though.
     
    Last edited: Aug 26, 2013
  2. jcsd
  3. Aug 26, 2013 #2
    Notice you have: sin2t + cos2t......
    That will simplify to 1 + 2sintcost / sintcost

    Try going from there..

    Are you converting to tan because the books answer is in tan form?? Swapping to Tan isnt necessary. If anything 1/sintcost can become csctsect
     
  4. Aug 26, 2013 #3

    haruspex

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    It's often not clear what should be regarded as the simplest form. If you define at as minimising the number of references to trig functions, you can get this down to one reference. Are you familiar with a formula involving sin(2x)?
     
  5. Aug 26, 2013 #4
    Oh duh! We didn't review any of the Pythagorean Identities in class, so I just assumed they wouldn't be used here. That, and I suck at trigonometry!

    Here it is again:

    (sin t + cos t)2
    /
    sin t cos t =


    1/(sin t cos t) =

    csc t sec t
     
  6. Aug 26, 2013 #5
    Well, probably not. I just know the Law of Sines, Law of Cosines, and Pythagorean Identities.

    But I do admit this forum is giving me a great "crash course" in trigonometry.
     
  7. Aug 26, 2013 #6
    You have to FOIL out the parenthesis. (sint + cost)2 does not equal 1
    You will have:

    sin2t + cos2t + 2sintcost / sintcost

    which gives you when simplified:

    1+ 2sintcost / sintcost

    Now, simplify.....
     
    Last edited: Aug 26, 2013
  8. Aug 26, 2013 #7
    Ok, I see what I did wrong (I took sin x + cos x = 1 when in reality it is sin2 x + cos2 x =1).

    I think I got it now (or I hope):

    1+ 2sin t cos t / sin t cos t

    [1/sin t cos t] + [ (2sin t cos t) / (sin t cos t) ]

    csc t sec t + 2
     
  9. Aug 26, 2013 #8

    haruspex

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    That's a right answer. To get it down to one trig reference you can use sin(2x) = 2 sin(x)cos(x).
     
  10. Aug 26, 2013 #9
    Thank you! It's a miracle I did it correctly, with the forum's help, of course. A lot of help, actually :)
     
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