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Trigonometric Identities

  1. Dec 3, 2009 #1
    hello.

    i have a question about trignometric identities.. it's realtivly easy, but am struggling with the algebra in it ( Algebra + trig = Very confusing to me )


    Prove that ..

    [Sinx/(1+Cosx)] + [(1+cosx) / sinx] = 2csc x



    i manged to get it to [Sin2x+1+2cosx+cos2x] / SinxCosx

    but i don't know what to do after that

    any help would be appreciated, even if hints or steps without actually solving it for me .

    Thank you
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 3, 2009 #2

    berkeman

    User Avatar

    Staff: Mentor

    There's an identity involving sin2x and cos2x that might help get you a bit farther...
     
  4. Dec 3, 2009 #3
    For the first term sinx/(1 + cosx), multiply everything by the conjugate of the denominator. It should be a lot simpler than what you tried doing. :smile:
     
  5. Dec 3, 2009 #4

    That should actually be [tex]\frac{sin^{2}x+1+2cosx+cos^{2}x}{sinx+sinxcosx}[/tex]

    Simplify and then factor.
     
  6. Dec 3, 2009 #5
    Woohoo got it !! ..

    Sin2x + cos2x will be 1

    so ..

    it will be [2 + 2cosx] / [sinx + sinxcosx]

    Cosx cancels each other

    ==> 4/ 2sinx

    ==> 2/sinx

    ==> 2cscx

    Thanks alot for the help guys :D
     
  7. Dec 3, 2009 #6
    I have anther question...

    Find the general solutions for the following equations.

    2 cos2 x -1 =0

    Soultion.

    Cos2x = 1/2
    Cosx= - or + Sqrt 2 / 2

    The general solution is pi /4 +2nPi, 3pi/4 +2nPi , 5Pi/4 + 2nPi and 7Pi/4 +2nPi , where n is an integer




    But why can't i just take the Double angle of 2cos2x ?

    which will be, cos2x=0

    The general solution is Pi/2 +2nPi and 3Pi/2 +2nPi where n is an integer..



    Btw, that was an example from my text book...the first solution is what the text book stated, and the second solution is what i thought right ..
     
  8. Dec 3, 2009 #7
    Mspike, don't make that mistake. The cosx don't divide out. Since you have already pretty much attained your answer, this is what should have happened:

    [tex]\frac{2+2cosx}{sinx+sinxcosx}[/tex]

    [tex]=\frac{2(1+cosx)}{sinx(1+cosx)}[/tex] The (1+cosx) in the numerator and denominator divide out.

    [tex]=\frac{2}{sinx}[/tex]

    [tex]=2cscx[/tex]
     
  9. Dec 3, 2009 #8
    Ahh!.
    thanks for the heads up, and i thought i knew it :S
     
  10. Dec 3, 2009 #9
    In your solution, you did not even mention pi/4. The double angle is as follows:
    2cos[tex]^{2}[/tex]x-1=0
    cos2x=0
    2x=pi/2
    x=pi/4

    That is only one solution, and you are still missing it. Check where you went wrong.
     
  11. Dec 3, 2009 #10
    Am sorry for asking so many questions ,

    how one should factor , tan2-sec-1=0

    i think it will be something like, (x-1)(x2+1)

    where x and x2 are trig
     
  12. Dec 3, 2009 #11

    rock.freak667

    User Avatar
    Homework Helper

    try using an identity involving tan2x and sec2x to replace the tan2x term.
     
  13. Dec 4, 2009 #12
    Solve Tan2x-secx-1=0

    Soultion

    (Sec2x-1)-secx-1=0

    sec2-secx-2=0


    (secx-2)(secx+1)=0


    --------------
    Secx -2 =0
    secx=2
    x= Pi/3 , 5Pi/3

    -------------
    Secx+1=0
    Secx=-1
    x= Pi


    X= Pi/3 , 5Pi/3, Pi


    Is that right? am i missing something ?

    Thank you :D
     
  14. Dec 4, 2009 #13

    rock.freak667

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    Homework Helper

    looks fine to me
     
  15. Dec 5, 2009 #14
    Yup, if the domain was [0,2pi], then it is correct.
     
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