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Trigonometric problem

  1. May 26, 2008 #1
    1. The problem statement, all variables and given/known data

    1)
    COS4[tex]\theta[/tex]-COS2[tex]\theta[/tex][tex]/SIN4\theta-SIN2\theta[/tex]=-TAN3[tex]\theta[/tex]

    2)sinx/cosx+1 + cosx-1/sinx = 0



    2. Relevant equations
    1) Verify
    2)verify

    3. The attempt at a solution
    1)
    cos(2[tex]\theta[/tex]-2[tex]\theta[/tex])-cos2[tex]\theta[/tex] / sin(2[tex]\theta[/tex]+sin2[tex]\theta[/tex])-sin2[tex]\theta[/tex]

    when simplified i get a large answer :S

    2)
    sinx/cosx+1 X cos-1/cos-1(reciprocal) + cos-1/sinx
    = sinx cosx-1/ cos2 -1 + cosx-1/sinx
    =sinx cosx-1/ Sin2x + cosx-1/sinx
    =cosx-1/sinx + cosx-1/sinx
    =2(cosx-1)/sinx :S


    thats it i hope u can read it
    formulas used
    Trigonometric Identities
    sum and difference Formulas of cosines and sines
    and double angle formulas

    my problem is that there is so many formulas and its hard to tell which one to use
    they are all usable but not all give u the answer
     
    Last edited: May 26, 2008
  2. jcsd
  3. May 26, 2008 #2

    rock.freak667

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    Homework Helper

    Find what

    Sin(A+B)-Sin(A-B) and similar for cos ,for the first part.

    [tex]\frac{sinx}{cosx+1}+\frac{cosx-1}{sinx}[/tex]

    Just bring them to the same denominator
     
  4. May 26, 2008 #3
    well how do u get them = to zero
    this far, and i don't know if its right :P
    cosx-1/sinx + cosx-1/sinx they gave the same denominator but they dotn' = zero
     
  5. May 26, 2008 #4

    rock.freak667

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    Homework Helper

    [tex]\frac{sinx}{cosx+1}+\frac{cosx-1}{sinx}[/tex]


    [tex]\frac{????}{(sinx)(cosx+1)}[/tex]


    bring them to a common denominator like that one.
     
  6. May 26, 2008 #5
    [tex]\frac{sin^{2}x+cos^{2}x-1}{(sinx)(cosx+1)}[/tex]


    ?
     
  7. May 26, 2008 #6

    rock.freak667

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    Homework Helper

    correct.

    What is [itex]sin^2x+cos^2x[/itex] equal to?
     
  8. May 26, 2008 #7
    :biggrin: 1
    thx man
    too simple and i didn't look that :P
     
  9. May 27, 2008 #8
    Are you still looking for help on this one?
    Hint: this is a very straightforward case of sum-to-product substitution
     
    Last edited: May 27, 2008
  10. May 27, 2008 #9
    2cos theta / 2sin theta

    when i use the double angle formula I end up with squared cosines and sines :S
     
  11. May 27, 2008 #10

    rock.freak667

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    Homework Helper

    Don't use the double angle formula here,it'll get too tedious

    Consider this
    sin(A+B)=sinAcosB+sinBcosA
    sin(A-B)=sinAcosB-sinBcosA

    if we add those two we get

    sin(A+B)+sin(A-B)=2sinAcosB

    Let P=A+B and Q=A-B, you'd eventually get A=(P+Q)/2 and B=(P-Q)/2

    hence then

    SinP+SinQ=2sin[(P+Q)/2]cos[(P-Q)/2]

    now do the same for

    cos(A+B)-cos(A-B)
     
  12. May 27, 2008 #11
    thx that helps
     
    Last edited: May 27, 2008
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